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AS Level Mathematics 9709 | Coordinate Geometry 8 | Standard Equation of Circle | Raja Shafqat — Transcript

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0:08Assalamu Alaikum, beta. In the last
0:10class, we looked at the introduction of
0:12the standard form. Okay? Today, we will
0:14finally practice the standard form more
0:16. Okay? We will practice the standard
0:18equation of a circle more. Now let's
0:20start with the standard equation of a
0:22circle. Number one. Let me give a quick
0:25recap to revise a little bit of what we
0:27studied last class. Then, Insha'Allah,
0:29we will continue further. Okay, beta?
0:31Then we will continue further. Last
0:33class, we learned something new. What
0:36is the standard equation of a circle?
0:39Okay? Last class, we learned something
0:42new. Normally, we hear about the
0:44equation of a line or the equation of a
0:46curve. So, in the last class, we
0:47learned something new. The standard
0:49equation of a circle. And what did we
0:51learn about the standard equation of a
0:53circle?(x-a) ² (y-b) ² Okay? We
0:59called this equation the standard
1:01equation of a circle = r squared. This
1:04is what we studied in the previous
1:05class, the standard equation of the
1:07circle. Clear, beta? This is what we
1:09studied in the last class. Okay? Now,
1:11finally, we are continuing with the
1:13same. Okay? Now, Insha'Allah, we will
1:16practice how this is applied. Okay?
1:19Also, we learned another thing: what
1:22are A and B in this equation? A and B
1:24are read as the coordinates of the
1:26center. A and B are the coordinates of
1:28the center. And then we learned what R
1:30is. R is the radius of the circle. And
1:34then we learned what X and Y are. Any
1:36point on the circumference. What are X
1:39and Y? Any point on the circumference.
1:41So, let me show you once again. We
1:44studied last time that this is your
1:46center, we named the coordinates of the
1:49center A, B, here it is. What was R? It
1:52was the radius. And what was P? P is
1:54basically any point on the
1:56circumference. Any point that is on the
1:58circumference, we call it P. Clear,
2:00beta? We call it P. Any point that will
2:02be on the circumference of the circle.
2:04Obviously, there are infinite points on
2:06the circumference of a circle. There
2:08are unlimited points. So, that was our
2:10lecture from the last class. Today too,
2:13we will only and only use the standard
2:15equation of a circle. But today, we
2:17will also add the properties of the
2:19circle. Okay? Let's start slowly. First
2:22of all, we will start with an average
2:24question. Question number three. Find
2:27the equation of the circle. Mention an
2:30equation of a circle whose center is
2:31this and passing through this point. So
2:34, if you have read this properly, what
2:36hint is given this time? Look, in this
2:39question, you already have the center
2:41of the circle. In this question, the
2:43examiner is already giving you the
2:45center of the circle. And the examiner
2:47is also giving you a point on the
2:48circumference. Mark another point
2:50anywhere. The point I am marking is a
2:52point on the circumference. Now, let's
2:55see what the coordinates of the center
2:57are. If you read this question, the
2:59coordinates of the center you have are
3:012 5. What are the coordinates of the
3:02center? 2 5. What is its radius? I
3:06don't know what the radius is. And a
3:09point is given, and the examiner is
3:11clearly saying that the circle passes
3:13through this point. Obviously, wherever
3:15this point is, its distance from the
3:17center will be equal to what? It will
3:19be equal to the radius. Right? Meaning,
3:21when you join the center of the circle
3:23to that point P, what will be formed? A
3:25radius will be formed. So, this is not
3:27an extraordinary question. We will name
3:29this point of yours point P. Now our
3:31target is the equation of this circle.
3:33We already know what the equation of a
3:35circle is. (x-a) ² + (y-b) ² = radius
3:40squared. This is the equation of the
3:42circle. Now only one thing is missing,
3:45what is the center of this circle? What
3:47is the radius of this circle? So, we
3:49just have to make a little extra effort
3:51to find the radius. If I find the
3:53radius, the radius will come with the
3:55help of the distance formula. The
3:57radius is obviously a length. The
4:00radius is obviously a length. So tell
4:02me, this becomes x1 y1 and this becomes
4:04x2 y2. x1 y1, x2 y2. So if I apply the
4:07distance formula, what is the formula
4:09for distance?(x2-x1) ² (x2-x1) ² and
4:15(y2-y1) ², here it is. When I simplify
4:20this, this will become 16 and this will
4:22become 9. What is the root of 16 and 9?
4:2416 + 9 is 25, and what is the root of
4:2725? 5. So, the slight extra effort here
4:30was that you had to find the radius
4:32first. What was the radius found for
4:34this question? The radius was found to
4:36be five. So now, if you want the
4:38equation of the circle, you already
4:40know a and b. You already know that the
4:42names of the coordinates of the center
4:45are a and b. So now, what has this
4:47question become? This question becomes
4:50x-2 squared + y-5² = radius squared,
4:53and what will the radius squared be?
4:57The radius squared will be 25. So today
4:59as well, we are not doing anything new.
5:01Today, we will use the equation of a
5:03circle in different situations. Clear,
5:06son? We will use the equation of the
5:08circle according to the question. Clear
5:11, son? This first one was just a very
5:13basic example. Look at another question
5:15of this same type. Then I will move on
5:16and have you do slightly tougher sums
5:18one by one. One by one. The diameter of
5:20the circle. The diameter of the circle
5:23has its endpoints at this and this.
5:25Find the equation of the circle. This
5:27is also almost the same kind of
5:28question. Well, a similar question has
5:30also appeared in past papers. The same.
5:32When I have you do past papers, I will
5:34show it to you again, God willing. Look
5:36, suppose you have a circle here. You
5:39have a circle here. Its diameter—
5:41which line is the diameter? In math,
5:43the diameter is called the line through
5:45the center. The line through the center
5:46. Now, are both endpoints of the
5:48diameter given to you? These two
5:50endpoints are given. Endpoints of the
5:52diameter. Obviously, we know that in
5:54math, the line through the center is
5:55called the diameter. In math, the line
5:57through the center is called the
5:59diameter. Isn't that right? Which line
6:01is the diameter? The line through the
6:03center. In the language of math, we
6:04call this the diameter. Okay. Now, I
6:07know where the center will be. The
6:09center will be its midpoint. Obviously,
6:12the midpoint of the two points of the
6:14diameter—what will that be? That will
6:16be the center of the circle. So, what
6:17are the endpoints of the diameter? The
6:20diameter's endpoint is -6 8. By the way
6:23, this is just a rough sketch. I am not
6:24doing any proper sketching. Clear, son?
6:27That I am showing -6 and 8 properly.
6:29Absolutely not. I am only giving you an
6:31idea. By the way. What are the
6:33coordinates of A? The coordinates of A
6:35are -6 8. The coordinates of A are -6 8
6:38. These are your coordinates for A. One
6:41endpoint is A. And its other endpoint,
6:43the second endpoint, what is it? 2 -4.
6:47This is your second endpoint. Now we
6:50know that when we talk about the
6:52equation of a circle, we require the
6:53center and the radius. Basically, two
6:55things are required. So first of all,
6:57its midpoint—the midpoint of the
6:59diameter, the midpoint of these two
7:02endpoints—who will that be? The
7:04center of the circle. So then, how will
7:06the center of the circle be found? We
7:08will say, how do we get the center of
7:09the circle? The center of the circle is
7:12the midpoint of AB, the midpoint of the
7:15line AB, or the midpoint of the line
7:17segment AB, it's all the same thing. So
7:20now, when you find its midpoint, the
7:22midpoint means the center. Coordinates
7:24of the center. Is it clear, dear?
7:27Midpoint means the coordinates of the
7:28center. So now tell me, what are these
7:31two points? -6 + 2/2 and similarly, add
7:34the y-coordinates of both. 8 + (-4)/2,
7:38so from here I found out what the
7:40coordinates of the center are. -4/2,
7:44that is -2, and 4/2, that is 2. These
7:47are the coordinates of the center you
7:49have found. Meaning one task has become
7:51easy for you. At least we have the
7:52coordinates of the center. Now you know
7:55what the coordinates of your center are
7:57. -2, 2, this is your center. Is it
8:01clear, dear? This is your center. Let
8:03me repeat it again. I haven't marked
8:05these as proper points. I just showed
8:07you a rough sketch. I showed you a
8:09rough sketch. Clear? Come on. Now we
8:12know how the radius will come. There
8:14are two or three options for the radius
8:16. Number one, radius will be the
8:18distance from A to C. Radius will be
8:20the distance from C to B. Radius will
8:22be the diameter length divided by 2, so
8:24there are two or three options to find
8:26the radius. Number one, the distance
8:29between A and C, this is also a radius.
8:31The distance between C and B, this is
8:33also a radius. Or the distance of AB/2,
8:36meaning find the length of AB and
8:38divide it by two. Find the distance of
8:41AB and divide it by two. You get the
8:42radius. Or just find the distance of
8:44this. You get the radius. Or just find
8:46the distance of this. You get the
8:48radius. Is it clear, dear? So these are
8:49multiple options. Suppose I am finding
8:51it out. I am applying the distance
8:53formula on AC. I am applying the
8:55distance formula on AC. Distance
8:57formula on AC, so look, what is this?
9:00X1, Y1, X2, Y2, I applied the distance
9:02formula. So this will be -2, this will
9:06be -2 + 6 squared, here it is, and + 2-
9:098 squared, here it is. Is it clear,
9:12dear? What did we simply do? We applied
9:14the distance formula. We applied the
9:17distance formula. So what did this
9:18become? 6-2 is 4. The square of 4 is 16
9:22. 2-8 is -6. The square of -6 is 36. So
9:26now, if I add both of these. If I add
9:29these two, what will be found from here
9:32? The radius will be found from here.
9:34So what is the answer to this? 52, this
9:36is your radius. If you apply this
9:38formula, distance between B and C, it
9:40will still be the same. 2-2, 4²-6²,
9:42indirectly it became the same thing. 16
9:46+ 36, so the radius is found. So
9:49finally, what is the radius answer? The
9:51radius answer is 52. The radius is 52,
9:55now the question has become easy. Now
9:58what is your target? Your target is the
10:00equation of the circle. This is the
10:02equation of the circle: (x-a) ² + (y-b
10:05) ² = radius squared. Meaning whenever
10:09the equation of a circle is required,
10:11you must have the coordinates of the
10:13center and the radius. Now tell me,
10:15what are a and b? Here is your a and b,
10:18a = -2, b = 2, now put these values. In
10:22place of a, -2, here you go. What
10:25should you put in place of b? Put 2 in
10:26place of b. Here you go. The radius
10:29squared, what will be the square of 52?
10:32The square of 52 will be 52. This is
10:34the standard equation of the circle
10:36derived for you, meaning the equation
10:39of the circle. This was your final
10:41target. Clear, son? This was your final
10:44target. So these two, both were average
10:46questions. These both were average
10:48questions. So today we will not do any
10:51new work. We will apply the standard
10:53equation of the circle in different
10:55situations and see how the equation is
10:57found. What thing will be missing? If
11:00any circle property is being applied,
11:02what property will it be? Clear, son?
11:05Come on, let's try more. Read the
11:07question properly. What is the examiner
11:10saying in this question? This is a
11:11practice exercise from the Cambridge
11:13book. So that the concept improves
11:15first. Then God willing, I will have
11:17you do its latest questions from past
11:19papers. Clear, son? I will have you do
11:21its latest questions from past papers.
11:23Come on, now read the question. What is
11:24the statement? Find the equation of the
11:27circle, that touches the x-axis, whose
11:29center is this. Touches the x-axis.
11:32Where will 6, -5 be? 1, 2, 3, 4, 5, 6,
11:351, 2, 3, 4, 5, it will be somewhere
11:38here, 6, 1, 6, -5. So, I am showing you
11:41a rough sketch, so that you first get
11:43an idea of what message the examiner is
11:45giving. Okay? 1 2 3 4 5 6, okay? Six,
11:51come on 1 2 3 4 5 6 1 2 3 4 5, let's
11:54suppose you have this situation, okay,
11:57now I will guide you a bit in detail
12:01about what the situation is. Okay? What
12:06is the situation? I am showing a rough
12:08sketch. I am showing it just so that
12:10you get a bit of an idea. So that you
12:12get a bit of an idea. The situation is
12:14something like this. Okay, son? The
12:16situation is somewhat like this. What
12:18is the examiner telling you? The
12:19examiner says that you already know the
12:21center of the circle. What is the
12:23center of the circle? The center of the
12:25circle is 6, -5. The coordinates of the
12:27circle's center are 6, -5. These are
12:30the coordinates of your center. Clear,
12:32son? Alright. Now, a little bit of
12:34common sense. Now, this is clearly
12:35given to us. Write this down. Find the
12:37equation of the circle that touches the
12:39x-axis. Touches the x-axis means this
12:42circle is tangent to the x-axis. What
12:44is this circle? By the way? This circle
12:46is tangent to the x-axis. This circle
12:48is tangent to the x-axis. I showed you
12:50a rough sketch. Now, if you know the
12:53properties of a circle, what does the
12:55radius always form? The radius always
12:58forms 90 ° with the tangent line.
13:01Clear, son? Meaning, this point here,
13:03if you remember, in mathematical
13:05language, we call it the point of
13:07tangency. What do we call this point in
13:10the language of math? Point of tangency
13:13. Meaning the point where your circle
13:16touches—any line that touches a
13:18circle at one point, we call that point
13:20the point of tangency. So, here we used
13:23a property that the radius is always
13:25perpendicular to the tangent line.
13:28Clear, son? The radius is always
13:30perpendicular to the tangent line. Now,
13:32coordinate geometry will be used a
13:34little bit. If you have a sense, what
13:36kind of line is this? This line is a
13:37vertical line. So we know that the
13:39equation of a vertical line is the same
13:41. x = constant. So on this entire
13:44vertical line, or let's say, if I ask
13:47you what the coordinates of this point
13:49of tangency are? Its y-coordinate is
13:52straightaway zero. On the x-axis, y is
13:55always zero. So what is its
13:57x-coordinate? Its x-coordinate is six.
13:59Why is it six? Because obviously, it's
14:02common sense 1 2 3 4 5 6, if your
14:04circle is something like this. This is
14:07its, this is its vertical distance.
14:09This is six. Okay? Suppose this is six
14:11for you. What does this technically
14:13mean? Technically, this means that the
14:16gap you see here on the x-axis, the
14:18equation of this line is right here.
14:21The equation of this line is right here
14:22. So, what is the vertical gap under
14:24the x-axis? The vertical gap under the
14:27x-axis is five. What does the minus
14:29sign show? The minus sign shows its
14:31locus, that it is under the x-axis. But
14:33length is always positive. So, this
14:35vertical distance—it's common sense
14:38that on a vertical line, there is
14:40always a change in y. Is that clear,
14:43kid? There is a change in y. So, what
14:45is this vertical gap? It is five. So,
14:47this length under the x-axis is five.
14:49And what is this length indirectly
14:51showing? This length is indirectly
14:52showing the radius. This length is
14:54indirectly showing the radius. So, what
14:56was your target? Your target is the
14:58standard equation of a circle. Your
15:00target is the standard equation of a
15:02circle. This is your target. Your
15:05target is the standard equation of a
15:07circle. So, what is the standard
15:08equation of a circle?(x-a) ² + (y-b)
15:13² = radius squared. So, tell me, what
15:17are a and b? a and b are 6 and -5. This
15:20is your a and b. So, what is a? Your a
15:23is 6. Here you go. What is your b? b is
15:25. Put in -5. Minus minus became plus.
15:28Radius squared. So, what will be 5
15:31squared? 5 squared will be 25. 5
15:34squared will be 25. So, what did we
15:37basically do here? Here we used a
15:39property of a circle. It was a small
15:41question. There is no very
15:42extraordinary property. So, here we
15:44used a circle property. Which property
15:47did we use? We used property number
15:49eight. What was property number eight?
15:52Property number eight was that the
15:55radius is always, the radius is always
15:57perpendicular to the tangent line. The
16:02radius is always perpendicular to the
16:06tangent line. Is that clear, kid? This
16:09is a small property we applied. So, see
16:12, coordinate geometry is a must, right?
16:13Obviously, this is also a part of
16:14coordinate geometry. So, on a vertical
16:16line, x does not change. So, if x is
16:18six here, then x is also six here. So,
16:20this gap—when we talk about length,
16:22the length is five. The minus sign with
16:25y indicates the location, that your
16:27point is under the x-axis. Its
16:29y-coordinate is negative. Meaning, the
16:31point is under the x-axis. Is that
16:33clear, kid? The point is under the
16:35x-axis. Done. Let's continue further.
16:39Okay? Let me continue further. Here,
16:41just read the language of the question.
16:43What is the statement? What is the
16:45examiner saying? Let's read the
16:46language first. Then I will guide you.
16:48Okay? Then I will guide you. What is
16:50the examiner saying, number one? The
16:52points P and Q are already given. Lie
16:54on the circumference of the circle.
16:56Okay? Show that, we have to prove
16:58something. My job is to prove or show
17:00something. Show that the center of the
17:03circle lies on this line. Meaning, the
17:06line through the center should be this.
17:08The line through the center should be
17:10this. The examiner is giving you an
17:12idea. Okay? The examiner is giving you
17:14an idea. Clear, kid? I am showing you a
17:16rough sketch like this. Okay? I am
17:18showing a rough sketch. Here you go. A
17:19rough sketch like this. I am using the
17:21word again and again. Rough sketch.
17:23Clear, kid? Rough sketch. 1 -2, 7 1,
17:26okay? Where would 1 -2 obviously be? It
17:29would be somewhere here. 1 -2 like this
17:31, and 7 1, 2, 3, 4, 5, 6, 7, 1 like
17:35this. So both these points are on the
17:37circumference. I am just giving an idea
17:39. I am repeating it again and again. I
17:40used the word 'idea'. Okay? It's a
17:43rough sketch. By the way. I just marked
17:44a point like this. Here it is. Suppose
17:46this point is 1 -2. And I just randomly
17:49marked a point. Just like that,
17:50randomly, this point is 7 1, here it is
17:52. I just randomly marked a point. Okay?
17:55You mark it somewhere else. Here you go
17:57. No issue. You mark it somewhere else.
17:59Here you go. No issue. It's just a
18:01rough sketch, right? By the way. It's a
18:03rough sketch. I don't even need
18:04accuracy in this. So what are the
18:06coordinates of point P? 1 -2, here it
18:08is. These are the coordinates of point
18:10P you have. These are your coordinates.
18:11Coordinates of point P and what are the
18:14coordinates of point Q? The coordinates
18:16of point Q are (7, 1). 7, 1, this is
18:19given in the question. Well, both these
18:22points are on the circumference of the
18:24circle. Done. Now, if I join both of
18:27these, then obviously a chord will be
18:29formed. Obviously, what will be formed?
18:31A chord will be formed. Clear, kid?
18:33Obviously, you will have a chord formed
18:35, a chord. Okay. What the center of
18:38this circle is, I don't know, it will
18:39be somewhere. Suppose I don't know what
18:41the center of this circle is. You can
18:43plot the point for the chord however
18:45and wherever you like. No issue. Is it
18:47clear, son? All right? Suppose your
18:49circle looks something like this. Here,
18:51I will just draw it this way. There is
18:53no restriction. This is just a rough
18:55sketch. Suppose this is your circle and
18:57this is its center. The center must be
18:58somewhere here. I do not know what the
19:00coordinates of the center are. I will
19:01guide you a bit more in a while. What
19:03is the center? I do not know. What is
19:05the question? Show that the center of
19:08the circle lies on this line. Now, we
19:11need a property of a circle here.
19:13Alright, son? Let me guide you now. I
19:15need a small property of a circle here
19:18so that I can help you solve the topic.
19:21So, first I need the circle's property.
19:24Which property do I need? Let's check
19:26the property first. We studied a topic
19:28before this: Geometrical Properties of
19:30a Circle. In the geometrical properties
19:33of a circle, we learned that suppose
19:35you have a chord. There is a chord, and
19:37this is your circle's center. We
19:39learned that whenever you drop a line
19:41from the center, that line always does
19:44what to the chord? It bisects it. The
19:47property's name was: A line through the
19:49center bisects the chord. All right? I
19:52will mark the property first. So, what
19:54was the name of the property? The
19:56property's name was: A line through the
19:57center. A line through the center
20:02bisects the chord, bisects the chord.
20:06That line which you drop from the
20:08center, what does it always do? It
20:10bisects the chord. It divides the chord
20:12into two equal halves. We studied the
20:14entire logic behind why it divides it.
20:16Because this is a radius and this is a
20:18radius, so this triangle is always
20:19isosceles. We have already studied this
20:22fact. Now look, I will do one thing. I
20:24will draw a line. Here it is, I have
20:26just drawn a line randomly. All right?
20:28I just drew a line randomly like this.
20:30All right? I have drawn this line. Come
20:33on, let me explain it again. I will
20:35show you again. Suppose I have drawn a
20:37line like this, randomly. Here, where
20:39is this line passing through? This line
20:41is passing through the center. It is
20:43passing through. Absolutely, it is
20:44passing through. All right? So, if I
20:46want, I can erase the remaining portion
20:48of it. Obviously, there is no issue.
20:50Let's read it again. Show that the
20:53center of the circle lies on this line.
20:55Let's check. We studied a property that
20:58a line through the center bisects the
21:00chord, right? We studied a property
21:02that a line through the center bisects
21:04the chord. Is that okay, son? We
21:06studied a property that any line
21:07passing through the center will do what
21:09to the chord? What will that line do to
21:11the chord? It will bisect it. What does
21:13bisect mean? Bisect means it will
21:15divide it into two equal parts. We have
21:18already studied this. Clear? So, this
21:20means what angle will it form? It will
21:22form a 90 ° angle. So, let me make it
21:25easy for you. Now, if you have sense,
21:27then who is this line? Basically, this
21:30line is the perpendicular bisector of
21:34the chord PQ. What is this line? By the
21:39way, the perpendicular bisector of the
21:41chord PQ. So, if we remember coordinate
21:43geometry, then here, obviously, this
21:45topic is coordinate geometry. So
21:47coordinate geometry is a must, right?
21:49So, the first thing we have to find out
21:51is the midpoint of PQ. First of all, we
21:55have to find out the midpoint of PQ.
22:00Midpoint of chord PQ. Okay, midpoint of
22:03chord PQ! Come on, so I will find the
22:05midpoint first, then I will explain it
22:06to you. Here is the midpoint of chord
22:08PQ. What is the midpoint formula?(x1 +
22:11x2)/2 and (y1 + y2)/2, here it is. So,
22:16what are the coordinates of the
22:18midpoint from this? 8/2 is 4 and this
22:20comes to -1/2. Here, so what did we
22:23find out? We found the coordinates of
22:25the midpoint. Coordinates of the
22:27midpoint. Is that okay, son? I am not
22:29calling it the center, by the way. Okay
22:31? So, make this correction first. I did
22:33not use the word center. I used the
22:35word coordinates of the midpoint. Here
22:38it is. So, whose midpoint is this? This
22:40is the midpoint of the chord. Whose
22:42midpoint is this? This is the midpoint
22:43of the chord. 4, -1/2. This is the
22:46midpoint of the chord. Clear, son?
22:48Midpoint of the chord. Now tell me one
22:50thing. I can find the gradient of chord
22:53PQ. Gradient of chord PQ, so if I talk
22:56about its gradients, then the gradient
22:58formula is y2-y1 over x2-x1. Here, so I
23:01applied the gradient formula 3/6, so
23:03what did the gradient of the chord PQ
23:05become? The gradient of this chord PQ
23:08line is 1/2, so if our coordinate
23:10geometry is strong, then what will be
23:12the gradient of the perpendicular
23:14bisector? What will be the gradient of
23:16the perpendicular bisector? The
23:18gradient of the perpendicular bisector
23:19will be -2, there you go, -2. The
23:22gradient of the perpendicular bisector
23:24will be -2, this is the gradient of the
23:27perpendicular bisector. Okay, now what
23:29is your target? Your target is
23:32basically the equation of the
23:33perpendicular bisector. Your target is
23:36basically the equation of the
23:38perpendicular bisector, that is your
23:40target. Your target is the equation of
23:43the perpendicular bisector, meaning you
23:46have to apply the property in today's
23:48topic. How? What is the perpendicular
23:51bisector itself? A perpendicular
23:53bisector is a line. A perpendicular
23:55bisector is a line, by the way. So,
23:57through which point is the
23:58perpendicular bisector passing? Now you
24:00know that this perpendicular bisector
24:02of yours is passing through the
24:03midpoint. We already have the
24:05coordinates of the midpoint. It is
24:06passing through this point. 4-1/2, here
24:09it is. These are the coordinates of the
24:11midpoint, the midpoint of its chord. By
24:13the way. It is the midpoint of the
24:14chord. By the way. And what was its
24:17gradient? I have already found the
24:19gradient. We studied the equation of a
24:23line in coordinate geometry: y-y 1 = m
24:27(x-x 1), there you go, y-y 1 = m (x-x 1
24:31), clear, son? There you go. Now what
24:35is your target? To find the equation of
24:37the line. Now if you want, you can make
24:39it easy by taking the LCM. 2y + 1/2,
24:41multiply the two over there.
24:44Simplification is not an issue. By the
24:46way. Make it easy. This became 2y. Okay
24:49? 2y = 2, what did we do a little bit?
24:51We have to simplify this. I already
24:53know the gradient of the perpendicular
24:55bisector. I already know the midpoint.
24:57Okay? I already know the midpoint.
25:00Clear, son? Now if you want, you can
25:02make it a little easier. This became 2y
25:04+ 1 = -4 x + 16, there you go.
25:07Rearrange it. Rearranged, 4x + 2y,
25:10shift the one over. So 16-1 will become
25:14what? 15, so this was your ultimate
25:17target. This is what you had to show in
25:19this entire question. So basically,
25:21what was used in this question? A
25:23circle property was used in this
25:25question. Which property? We studied
25:28the property that if you have a circle
25:29and any chord, then when you drop a
25:31line from the center, it will bisect
25:33the chord. Now, you take the chord like
25:35this, or take it like that. Whenever a
25:37line is dropped from the center, it
25:38will always bisect the chord. Whenever
25:40a line is dropped from the center, it
25:42will always bisect the chord. Okay, son
25:45? A line through the center bisects the
25:47chord. A line through the center
25:49bisects the chord. Okay, son? We have
25:51already studied this. A line through
25:54the center bisects the chord. Okay,
25:55dear? Now let's continue further. Let's
25:58continue further. Come on. So the
26:01properties will apply. This is a must.
26:04Here you go. Now this question. Well, a
26:06question of the same style is in your
26:08past papers too. Exactly the same style
26:10. You will read it first, then I will
26:12guide you. The examiner says that
26:14points P, Q, and R are joined to form a
26:16triangle. Right? Which triangle? How
26:19would we know? Is it right-angled,
26:21equilateral, isosceles, scalene, acute,
26:24or obtuse? We don't know. Look, what is
26:27written here? Show that angle PQR is a
26:29right angle. Angle PQR. Right? Right?
26:33So if I read the question, the examiner
26:35is telling me that angle PQR is a right
26:37angle. I have to show, I have to show
26:39that the angle at point Q is 90 °. The
26:43angle at point Q is 90 °. Right, dear?
26:48So my first task is to show that it is
26:50a right-angled triangle. Look, what is
26:51written here? Triangle. So if any one
26:54angle in a triangle is 90 °, it is
26:56called a right-angled triangle. So what
26:58does that mean? That I have to prove a
27:00right-angled triangle, by the way. And
27:02at which point do I have to prove it?
27:03At point Q. Look, I am showing you a
27:05rough sketch like this. At which point
27:07do I have to prove it? At point Q. Now
27:09you can plot the triangle however you
27:12want. There is no issue. Clear? Plot it
27:14however you like. No issue. Right? Mark
27:17Q anywhere. Mark P anywhere. Is this an
27:20...accurate diagram? Right, dear? Not
27:23an accurate diagram. Mark P anywhere.
27:25Mark Q anywhere. Whatever you feel like
27:27. Plot it this way. No issue. Right,
27:30dear? There are no restrictions. What
27:32do you just have to do? You have to
27:34show that at a particular point, the
27:35angle is what? 90 °. So, plot the
27:37triangle however you want. So, where is
27:40your point P? Here is your point;
27:42suppose this is your point P. -5 6;
27:44what is your point Q? -3 8; here it is.
27:47Obviously, its X...its Y coordinate is
27:496. Its Y coordinate is 8. So, obviously
27:51, it will be higher up. I am just
27:52giving you an idea, by the way, an idea
27:54. Okay? And who is the third point? The
27:56third point is R. This point is 3 2,
27:58here you go. This point is 3 2, here
28:00you go. Read the question above. Is
28:02that okay, kid? Read the question above
28:04properly. What is the examiner saying?
28:06Show that angle PQR is a right angle.
28:08We have to show that this angle here is
28:1190 °. So, what is the indirect message
28:14here? That is, prove the triangle is a
28:16right-angled triangle. Prove the
28:18triangle is a right-angled triangle.
28:20That is the indirect meaning. So now,
28:22tell me one thing. I first applied the
28:24gradient formula. We know it. We have
28:26already studied this in coordinate
28:27geometry. Now, if I ask you what is the
28:29gradient of PQ? Then you would say y2-
28:33y1 over x2-x1, here is y2-y1 over x2-x1
28:35, so what result is derived? The result
28:38derived will be 2/2, so its gradient is
28:41one. Here you go, I have found its
28:42gradient. The gradient of the line is
28:44one. Alright, now I am finding. Now I
28:48am finding the gradient. Whose gradient
28:51? Now I am finding the gradient of QR.
28:53Here it is. We know the formula for the
28:56gradient of QR, right? y2-y1 over x2-x1
28:59, y2-y1 over x2-x1, so what is the
29:02formula? y2-y1 over x2-x1, here it is,
29:07so what is the answer? 6/-6 and what
29:10did the gradient come out to be? -1, 6/
29:126/-6, what did the gradients come out
29:15to be? The gradient came out to be -1,
29:17that is your gradient. So tell me, what
29:20is the product of both these gradients?
29:22-1, we know whenever we talk about
29:24perpendicular lines, we studied a
29:26condition in coordinate geometry that
29:28whenever lines are perpendicular, what
29:30is the product of their gradients? -1,
29:33okay? Meaning the gradients are always
29:36negative reciprocals of each other.
29:37What are gradients always? They are
29:39negative reciprocals. Is that clear,
29:41kid? So now, tell me one thing. So what
29:43will I show the examiner? That I will
29:45show the examiner. It will be the
29:47gradient of PQ multiplied by the
29:49gradient of QR, so what is the result
29:52derived? The result derived is -1,
29:54meaning what is the product of the
29:56gradients? What do you have for the
29:59product of the gradients? The product
30:01of the gradients you have is exactly -1
30:03. And this is what needs to be shown
30:05for perpendicular lines. Is that clear,
30:06kid? This is what needs to be shown for
30:08perpendicular lines. Okay? That was
30:10easy. The next part is interesting.
30:13Find the equation. Find the equation of
30:17the circle that passes through PQR, now
30:19this is a very interesting question.
30:22Okay? This is a very interesting
30:24question. Now, come look here and check
30:26a point. This is a somewhat interesting
30:28question. Clear? Check the point. So,
30:31what did the examiner do? See how the
30:33examiner connected the question? A
30:36little while ago, the examiner had you
30:38prove something. Two minutes ago. Look,
30:41I am showing you a rough sketch like
30:42this. Two minutes ago, the examiner
30:44asked you to prove that angle PQR is a
30:46right angle. Two minutes ago, the
30:47examiner asked you to prove that this
30:49triangle is a right triangle. Prove
30:51that this triangle is a right triangle.
30:53Right? Now look, I will connect a
30:56property to this particular topic for
30:58you. Okay? So that you can develop a
31:02sense of what the examiner has done.
31:04Okay, son? Develop a sense of what the
31:07examiner has done. Clear? Well, there
31:09was a property called the semi-circle
31:12property. Semi-circle property. Okay?
31:16If any student remembers geometrical
31:17properties, they will immediately say
31:19there was a semi-circle property. The
31:21examiner basically did this. The
31:24examiner basically did this. Okay? I
31:26will guide you. Okay? Just like the
31:30sketch above. If you want, take this
31:31same triangle and fit it inside this
31:33circle. Okay? This is the same triangle
31:35I just found. Let me show you by
31:38fitting this same triangle inside it.
31:40Okay? The same triangle, because both
31:42these parts are connected to each other
31:44. Look at this. So, what did the
31:46examiner do? I will enlarge it a little
31:48to show you. The examiner had you prove
31:51that this angle here is 90 °. Here you
31:53go. So, what did I do? I took that same
31:55triangle and copied it here. Okay? Here
31:58you go. Come on, take this. Copy the
32:00same triangle here. Here you go. Take
32:02it from there and copy it here. Now,
32:04what does this mean? We had studied a
32:07property. Let me guide you again on
32:09which property is being used in this
32:11question. A property is being used. I
32:13will mark the property here. The name
32:16of the property was the semi-circle
32:18property. We studied 10 properties of a
32:21circle, and that property was called
32:22the semi-circle property. What was the
32:24semi-circle property? We studied the
32:27property that the angle subtended by a
32:29diameter is always 90 °. Whenever both
32:33points of a diameter join at the
32:35circumference, they always form 90 °.
32:38What do we call the line through the
32:40center in math? We call the line
32:42through the center the diameter in math
32:44. What do we call the line through the
32:45center in math? We call the line
32:47through the center the diameter in math
32:48. Is that clear, son? The line through
32:50the center is named the diameter. Okay?
32:53The line through the center is named
32:54the diameter. All right. So, the line
32:57opposite to 90 ° is the diameter,
32:59isn't it? So, what does that mean this
33:00line is? This line is your diameter.
33:02This line is the diameter. All right?
33:05So, the name of this property was the
33:07semi-circle property. This means your
33:09midpoint will be on this line itself.
33:12Is it clear, son? This is your midpoint
33:14. So, I will write here to explain to
33:17you that PR is the diameter. PR is the
33:21diameter. Is it clear, son? PR is the
33:24diameter. Now, what is your target?
33:27Your target is the equation of the
33:28circle. Read the language. Equation of
33:30the circle. Equation of the circle.
33:32This is your target. So, first of all,
33:34you need the coordinates of the center.
33:37The name of these center coordinates is
33:39C. So, first you need C. Who is C? What
33:42is C? C is the center of the circle.
33:44What is C? By the way? C is the center
33:47of the circle. C is the center of the
33:51circle. All right. So, first find its
33:53midpoint. Because it is confirmed,
33:55right? This is the diameter. So, the
33:56midpoint of the diameter is the center.
33:58Obviously, obviously, so what did you
34:01do? You added both and divided by two.
34:03Similarly, you added these two as well
34:05and divided by what? Divided by two.
34:07Divided by two. Now, when we simplify
34:10this, it will be -2/2 which is -1 and
34:126/2, what will that be? Four. You have
34:15derived the coordinates of the center.
34:17So, first of all, the coordinates of
34:19the center have been found. So, look at
34:22how it connected. The right-angle
34:23triangle with the semi-circle property.
34:25Is it clear, son? We had to see how it
34:27connected the diameter to the
34:29semi-circle property. Now look at one
34:32more thing, son, look at one more thing
34:34. Now, if I ask you what is the radius?
34:38Radius. So, the gap from the center to
34:41the circumference, the gap from the
34:44center to Q. Look here, the gap from Q
34:46to the center. Clear? These are all
34:49radii, obviously. All right? If I ask
34:51you, what is its radius? So a student
34:53might say, sir, any gap you take from
34:54the center to the circumference is the
34:56radius. Look, this is also a radius.
34:59This is also a radius. This is also a
35:01radius. What are all these lengths? All
35:04these lengths are the radius. What is
35:05this one as well? It is the radius. So,
35:07if I ask you that I need the radius of
35:09this. I require its radius. Then you
35:11have many options. How to get the
35:13radius, i.e., r? The distance between P
35:16and C is also the radius. Okay? Okay?
35:18Similarly, the distance between Q and C
35:20is also the radius. Similarly, the
35:22distance between R and C is also the
35:23radius. In the same way, I can also say
35:25that take the length of PR and divide
35:27it by two. This is also the radius.
35:29What are all these? By the way, all of
35:31these are the radius for you. All of
35:33these are radii. So, you can do any one
35:35of these. The coordinates of the center
35:37have been found, right? What are the
35:38coordinates of the center? -1, 4, so
35:40let's suppose I find out the length of
35:43PC. That is, I applied the distance
35:46formula to this. I am basically finding
35:48this out. Here is the radius. You can
35:50do whichever one feels easy to you. By
35:52the way. So I applied the distance
35:54formula. So this will be -1-5, that is
35:58(x2-x1) ², then it becomes + 4², next
36:03(y2-y1) ² is 4-(-2) ², there you go.
36:08Okay? So what is the answer to this? 16
36:10+ 4 = 20, 16 + 4 = 20, this is your
36:13radius. What is the radius? 20, if
36:16anyone doubts my working, use these two
36:18points. 3-(-1) = 4², -2², check it,
36:2320, same result, right? This means that
36:26whatever you do, what will the result
36:28ultimately be? The result will
36:30ultimately be the same. What was your
36:31target? The radius. The radius has
36:33arrived. Finally, what did we need? The
36:36equation of the circle. Equation of the
36:38circle. This is our target. What
36:41equation of the circle did we study?(
36:43x-a) ² + (y-b) ² = radius squared.
36:48This was the equation of the circle.
36:50Now tell me, what are the coordinates
36:51of the center? a is -1. What is b? b is
36:55your 4, because we call the coordinates
36:57of the center a and b. This is your a.
36:59This is your b. Your a is -1 and b is 4
37:04. Put it in. Equals radius squared. So
37:07what will be the square of root 20? The
37:09square of root 20 will be 20. What did
37:11we find out here? We have found the
37:15equation of the circle. The equation of
37:18the circle has been found. Is that
37:20clear, son? The equation of the circle
37:22has been found. This was basically our
37:25target in this question. Clear? So,
37:28look today. I showed you four questions
37:30. But what did we use in all of them?
37:32In all of them, we used circle
37:34properties. That is why we worked in a
37:36sequence. First, we studied quadratics.
37:38Then we studied circular measurement.
37:40Because the concepts involved in both
37:42are used in this topic. Clear, son? The
37:45concepts involved in both are used in
37:47this topic. Done? So finally, we
37:50haven't done anything new yet. Okay? We
37:53haven't done any new work. We are just
37:55doing the same work we studied before.
37:57Clear, son? Let's continue further now.
38:00Okay? Let's continue now. Here you go.
38:03Read the question, please. I am waiting
38:04for you. Please read the language
38:06carefully. Language is the most
38:08important thing; first, you must read
38:10the circle question properly to see
38:12what hint the examiner is giving. Okay?
38:15Because if you don't read it properly,
38:16there will be no understanding, and it
38:18will be impossible to solve the
38:19question. Let's find the equation of
38:22the circle. We know what the equation
38:24of a circle is, right?(x-a) ² + (y-b)
38:27² = radius squared. This is the
38:30equation of the circle that passes
38:31through this point and this point. Is
38:33that right? And has its center lying on
38:35this line. The center lies on this line
38:37. I am not saying this. The examiner is
38:39saying this. Okay? Now I will show you
38:41what the situation is, by the way.
38:43Meaning, what is the indirect message?
38:45Or what does this question technically
38:47mean? Okay? Alright, one point is 7, 3;
38:511, 2, 3, 4, 5, 6, 7, 3, somewhere
38:54around here is 7, 3 and one point is 11
38:57, -1; 7, 8, 9, 10, 11, -1, meaning it
39:00is like this. One point is somewhere
39:02here, one point is somewhere here; I am
39:03just giving you an idea. I am using the
39:05word 'idea' repeatedly. Okay? So it
39:07means both these points are on the
39:09circumference. I am just giving you an
39:11idea. I am using the word repeatedly.
39:12Idea. Okay? Here it is. One point is
39:15here. One point is here. Let's name one
39:18point P. One point is 7, 3, there you
39:21go. This point is 7, 3. We named it P.
39:25And there is one more point, its
39:26coordinates are 11, -1, here it is. 11,
39:29-1. This is another point given, 11, -1
39:31. All right? All right? This is another
39:34point given. We have read this. What is
39:36the center of this circle? Suppose I
39:38don't know. What is it? It could be
39:40anything, I don't know. What is the
39:41center of the circle? All right? Let's
39:43suppose this is the center of the
39:44circle. We know that the center of the
39:45circle is called C and its coordinates
39:47are called a and b. Its coordinates are
39:49called a and b. Done. Wait, what is
39:52written? Okay. And has its center on
39:54this line. The examiner is saying that
39:57the center of this lies on this line.
40:01All right? The examiner told me that
40:03its center lies on this line. Like this
40:07, the center of the circle lies on this
40:09line. All right? The center of the
40:13circle lies on this line. I haven't
40:15solved anything yet, by the way. Just
40:17given an idea, only an idea. All right?
40:20So that a little sense develops as to
40:21what the examiner wants to do. Clear?
40:24What the examiner wants to do. Just so
40:26a little sense develops. So that a
40:28little sense develops. All right? Here
40:30it is. What is the center? I don't know
40:32the center. Suppose what is the center?
40:34I don't know. But we do know what a
40:35center is, right? The center has
40:38coordinates a, b; a, b is your center.
40:41The center has coordinates a, b. A
40:45circle property will be used. We will
40:47have to apply a property here again. I
40:49will guide you again. First, a property
40:51of the circle will be used. First, I
40:53will repeat that circle property. Then
40:55I will show you the question. So this
40:57means that in all these questions,
40:59circle properties are a must. Now tell
41:01me one thing. We studied a property
41:03before as well that a line through the
41:05center bisects the chord. We studied
41:07that when we drop a line from the
41:09center, what does it always do? It
41:12bisects the chord. A line through the
41:15center bisects the chord. Isn't that
41:17right? A line through the center
41:19bisects the chord. So this same
41:20property is being used indirectly again
41:23. A line through the center bisects the
41:25chord. We need this property. Then I
41:26will guide you on how we need it. Line
41:30through the center. This property is
41:33called line through the center. A line
41:36through the center bisects the chord. I
41:39am writing down this property for you
41:42so that when you revise or if you don't
41:44remember it, you will be reminded. Now
41:47tell me one thing. Now I know that the
41:50line that bisects these chords, look, I
41:52will draw a white line and show you.
41:54Okay? I will draw a white line for you.
41:57This line that bisects the chord will
41:58also pass through the center. So, this
42:01white line I have drawn, look again, I
42:04have bisected this chord. Look, I have
42:07bisected this chord. So, we already
42:09know, don't we, that a line through the
42:11center will do what? A line through the
42:13center bisects the chord. When I bisect
42:15this chord, will this line also pass
42:17through somewhere? This line will also
42:19pass through the center. So, I applied
42:20a property: a line through the center
42:22bisects the chord. This means, who is
42:25PQ? PQ, who was PQ? PQ, by the way, was
42:28the chord. Who was PQ? PQ is your chord
42:32of a circle. And this white line, who
42:35is it? It is the perpendicular bisector
42:37of the chord. So, what will I call this
42:39white line? This white line is the
42:42perpendicular bisector of the chord.
42:45What is this? The perpendicular
42:46bisector of the chord. Can you see what
42:48I did? I bisected this chord. So, we
42:51know the property, right, that whenever
42:53any line bisects a chord, it passes
42:54through the center. It passes through
42:57the center. Confirmed. And this line is
42:59already given to us. Now, what is
43:01missing for us? What is our ultimate
43:04target? Our target is the equation of
43:05the circle. For a circle, we need two
43:07things. The center and the radius. Both
43:10the center and the radius. Number one.
43:13Well, the first thing is that this line
43:14is already given to you. What line is
43:16given? By the way, you have this line
43:18given? 2x + y = 7, here it is, 2x + y =
43:237, this is already given. By the way, I
43:27will write it below; the given line is
43:28right here. I didn't do anything; this
43:30is already given. By the way, what is
43:32my target now? My target is the
43:36equation of the perpendicular bisector
43:39of chord PQ, because once that comes,
43:43my system will move forward. Then I
43:47will guide you again on what to do.
43:49Look, the first thing is that you have
43:51to find the equation of the
43:53perpendicular bisector. So we know what
43:55a perpendicular bisector does, right?
43:56It bisects a given line. To bisect,
43:59what do you need first? First, you will
44:02need the midpoint of PQ. The midpoint
44:05of PQ, so your first task is the
44:07midpoint. This is your very first task,
44:10the midpoint. Here you go. So how do we
44:11get the midpoint? X1 + X2/2, here it is
44:15. And the midpoint will be y1 + y2/2,
44:18here it is. So when I found its
44:21midpoint, the coordinates of the
44:24midpoint came out to be 18/2, meaning 9
44:27, and 2/2, meaning 1. 2/2 means 1. 2/2
44:30means 1. Clear, kid? 2/2 means one.
44:33Okay, the midpoint is found. So I have
44:35found this point. Right? I'll copy it
44:38here to explain to you that this point
44:40is 9, 1. There, this point 9, 1 is
44:42found. Secondly, we know what the
44:45gradient formula is, right? So first, I
44:47will find the gradient of PQ. We know
44:49the gradient formula is y2-y1 over y2-
44:52y1 over x2-x1, here it is. So this
44:56becomes 4/-4, meaning its gradient is
44:58-1. Whose gradient is this? This is the
45:01gradient of PQ. This is the gradient of
45:04PQ. Clear? So the gradient of this line
45:06, PQ, is -1. Here you go. So then, if I
45:09ask you what the gradient of the
45:10perpendicular bisector will be? What
45:12will be the gradient of the
45:14perpendicular bisector? It will be +1,
45:16that is, the negative reciprocal of
45:19this. Think, if this gradient is -1,
45:21then what should this be? It should be
45:23+1. Now look at its equation. The
45:26equation will be y-y 1 = m times x-x 1,
45:30here it is. Now you know the
45:31perpendicular bisector passes through
45:33the midpoint. Its gradient is found.
45:35Now simplify this. So y = x-9 + 1, here
45:39it is. So what is the answer from here?
45:42y = x-8, this is the equation of the
45:45perpendicular bisector you have.
45:48Equation of the perpendicular bisector
45:50of PQ, this is also the equation of the
45:52perpendicular bisector of PQ. Here it
45:54is, the equation of the perpendicular
45:56bisector is found. Done. Now these two
45:58lines are intersecting. Tell me one
46:01thing. When this line and this line
46:03both intersect, what will we get? The
46:05coordinates of the center. The
46:06coordinates of the center. A, B will
46:09come. Basically, the point that comes,
46:11that point will be named (A, B). Right?
46:14A, b, now tell me, are there two lines?
46:16Yes, there are two lines. What is the
46:17equation of one line? Look closely. One
46:21line's equation is 2x + y = 7, here is
46:24one line and another just came, y = x-8
46:27, these are the two lines. Now we will
46:31solve these two lines. Now what do we
46:33need to do with these two? We have to
46:35solve these two simultaneously so that
46:37we get C. What will we do with them? We
46:40will solve these two simultaneously for
46:43C so that the coordinates of the center
46:45are obtained. Replace y with y. Replace
46:49y with y. So this becomes 2x + x-8 = 7,
46:53this becomes 3x = 15, there it is. So
46:57what is the answer for x from here? The
46:59answer for x is 5. The answer for x is
47:025. So when the answer for x is 5, what
47:04is the answer for y? The answer for y
47:07is -3, there it is. So that means what
47:09are the coordinates of C? The
47:11coordinates of C are 5, -3, there you
47:13go, the coordinates of C, meaning the
47:15center of the circle, have finally been
47:18found. Finally, the center of the
47:20circle is here. Now only one thing is
47:22missing, which is the radius. Now tell
47:24me one thing. You have found the center
47:27of the circle. Okay, kid? We just found
47:29out the center of the circle. I have
47:31the coordinates of the center of the
47:33circle as 5, -3. So tell me one thing,
47:36point P is also on the circle, done.
47:37Point Q is also on the circle, done.
47:39These are all points on the circle,
47:40right? This is also on the circle, this
47:42is also on the circle, and C is also on
47:43the circle. I need the radius, so how
47:45will I get the radius? The distance
47:47between Q and C will be the same as the
47:49distance between P and C. This will
47:50also be the radius. This will also be
47:52the radius. Obviously, this will also
47:54be the radius. So this is also R for
47:56you. This is also R for you. Now you
47:58can find the radius using any two
48:00points. You already have P as 7, 3, P
48:03is already with you as 7, 3, let me
48:06repeat. Now your target is the radius.
48:08So what are the coordinates of P? The
48:10coordinates of P are 7, 3 and what are
48:13the coordinates of C? The coordinates
48:16of C you have are 5, -3. Here it is.
48:18What is your target? By the way, your
48:20target is the radius. So now I have
48:22applied the distance formula. Meaning I
48:25need the distance between P and C 5-7²
48:28+ -6² there it is, so this will be 40,
48:31so what did its radius come to? Its
48:34radius came out to be 40, so the radius
48:37of this circle is 40. Look, son, if any
48:40student has any doubt about what I said
48:43, they can definitely use this second
48:46point. Check the distance between these
48:48two points, Q and C. What was your C? 5
48:51and -3, 11-5 is 6², plus 2 squared,
48:54under root 40, it comes out the same.
48:58What is your target? Your target is the
49:00equation of the circle. What is your
49:02target? Equation of the circle. This is
49:05finally your target in this question.
49:07So, we know what the equation of a
49:09circle is, right? The formula for the
49:12equation of a circle is (x-a) ² + (y-b
49:15) ² = radius squared. This was your
49:18target finally. (x-a) ² + (y-b) ² =
49:21radius squared, you know this was
49:24finally your target in this question.
49:28Clear, son? This was your target. Now
49:30what do you have to do? Just say it. Do
49:32you know the center? Do you know the
49:33radius? What is the center? Tell me.
49:35The center is right here, 5 and -3.
49:37What do we call the center? We call the
49:39center a and b. a and b are right here.
49:41So now tell me, what will be its answer
49:42? a and b, so what will its answer be?
49:45Look, its answer will be (x-5) ². (y +
49:503) ². = radius squared, so what did
49:54the radius squared come out to be? The
49:56radius squared finally came out to be
49:5840, your target is achieved. Your
50:00target in this question is achieved. So
50:03what was our target? Finally, the
50:04equation of the circle was the target.
50:07Finally, as soon as we hear the word
50:09circle, it clicks immediately that, hey
50:11, this is the equation of a circle. So
50:14for a circle, there should be a center
50:17(a, b) and a radius r. This is the
50:20center of the circle, for which we had
50:22to put in a little effort. As soon as
50:25the center is found, the gap from the
50:27center to Q or the gap from the center
50:29to P, both are the radius; so you can
50:31do whatever you like, the radius will
50:33come out the same, the radius will come
50:35out the same, clear? So what is being
50:38used in all these questions? In all
50:39these questions, some property of the
50:41circle is being used. Meaning your
50:43questions are dependent on circle
50:45properties, on circle properties. Come
50:48on, let's try one or two more questions
50:50. Okay? Let's try one or two more
50:51questions of this type. Come on, now
50:53read the question and see what the
50:55examiner is asking us. A circle has a
50:58radius of 10 units and passes through
51:00this point. The x-axis is a tangent to
51:03the circle. Find the possible equations
51:07. It is plural here. Is that okay?
51:10First, it's plural. Now look at what
51:13it's telling you. It says the radius is
51:1510. Done. Alright? The radius is 10. It
51:18passes through 5 -6, we just need some
51:20common sense. 1 2 3 4 5, 1 2 3 4 5 6 7
51:238 9 10 11 12 13 14 15 16, here it is.
51:28It'll be somewhere here, 5, 16; radius
51:30is 10, so if the radius is 10, then it
51:33will go up to 6. This means your circle
51:36could look something like this. A
51:38circle could look something like this.
51:41Note this down for a minute. The circle
51:42will be like this. Let's understand the
51:44point first. The exact condition isn't
51:46given yet. First, let's read because
51:48look, 5 -16 is quite far. Here is 5 -16
51:51. The circle is passing through this
51:54point. Its radius is 10. The radius is
51:5610, so if this whole gap is 16, then 10
51:59will be less, right? 10 would be
52:01somewhere here, let's say. Is that
52:02clear? 10 up to here, meaning the
52:04radius is 10. Alright? Come on, look at
52:07it again. Let's say this is -16. Let's
52:10say this is -10. Its radius is 10. Its
52:14radius is 10. Okay? The situation is
52:16given something like this. So this
52:18could mean that your circle is
52:20basically something like this. Like
52:22this. Alright? Now the situation could
52:24be such that what is its radius? It is
52:2610. What is its radius? It is 10 and
52:29the circle is passing through another
52:30point. Radius length is obviously
52:32positive, so I will take it as positive
52:33. 5 -16 will be somewhere below that.
52:37It will be somewhere below that. Its
52:40radius is 10. Alright? Look closely for
52:42a minute. The radius is 10. And it's
52:44written that the x-axis is a tangent.
52:46Meaning it's touching the x-axis. This
52:49hint is already given. It is touching
52:51the x-axis. Okay, son? The situation is
52:55something like this. Even I don't know
52:57exactly right now. I just know that the
52:59situation is basically something like
53:01this. The situation is basically
53:03something like this, clear. Now if we
53:05think about it in more detail, then God
53:07willing, it will click what the
53:08examiner wants us to do, what the
53:10examiner wants us to do, done. Well,
53:12after a while, when it's solved, it
53:14will become more obvious. Let me
53:16explain it to you once again. Now let's
53:18start even slower. Find the possible
53:20equations of the circle. From "equation
53:22of circle," it immediately clicks that
53:24yes, this is what is called the
53:25equation of a circle in math. x-a²
53:29y-b² = r², so we need a and b, which
53:32is the center, and a radius. We did
53:35almost the same work in the previous
53:37question as well. Alright, first let me
53:39give you an idea. Let me just give you
53:43an idea. The examiner said that the
53:45x-axis is a tangent to the circle,
53:46right? Now, I don't even know if the
53:48situation is like this. Okay? I don't
53:50even know what type of situation it is.
53:52Maybe the situation is something like
53:54this for you. Okay? I don't know
53:56exactly, do I? It is possible that the
53:58situation is something like this. It is
53:59possible that 5, 16, -16 is somewhere
54:01around here. It is possible that it is
54:03somewhere here. How would I know
54:04exactly? I don't know exactly. Okay? I
54:06don't know exactly. After a while, we
54:09will know for sure. After a while, we
54:11will make a correction about what the
54:12exact situation is. What the exact,
54:14exact situation is. For now, I'm just
54:17showing you a rough sketch. Then, after
54:19a while, we will see it properly and in
54:21detail. After a while, it is confirmed
54:24that the x-axis is a tangent. I am not
54:27saying this. The examiner is saying
54:28this. So, first of all, if I don't know
54:30the center of the circle, mark a point
54:32somewhere like this. This is the center
54:35of the circle for you. What is its
54:37radius? Its radius is 10. Well, we
54:39already know this, right? We already
54:42know that the radius always forms 90 °
54:44with the tangent line. So the examiner
54:47says that the x-axis is a tangent. I am
54:49not saying this. The examiner says that
54:52the x-axis is a tangent to the circle.
54:55Meaning your circle is touching the
54:56x-axis only at one point, by the way.
54:59And okay, this became your radius. How
55:01much is your radius? Your radius is 10.
55:02Here it is. This is your radius, by the
55:04way. Okay? This is your radius. Well,
55:06now you are given another point.
55:08Obviously, you are given one more point
55:10. Where is that point? I don't know.
55:12That point could be anywhere around
55:13here. It is on the circle. But one
55:15thing I know, it will be below this. By
55:17the way. It will be below this, right?
55:19It is possible it's here. It is
55:20possible it's here. How do I know where
55:22it is exactly? I am just giving an idea
55:24. Okay? After a while, there will be
55:26confirmation, after a while. So, what
55:27are the coordinates of this given point
55:30? The coordinates of this point are 5,
55:32-16. This is the situation you have.
55:34This point is on the curve. I just
55:35showed you a rough sketch. A rough
55:38sketch. Done. Now, what is our target?
55:41Our target is the equation of the
55:42circle. The equation of the circle.
55:44Answer me one thing. The center of this
55:46circle will be a, b; that's obvious. If
55:49this line is perpendicular, then we
55:51already know what its coordinates will
55:53be, right? a, 0. Why? Because on a
55:55vertical line, x is always constant.
55:58Because when the radius is
55:59perpendicular to the tangent, it
56:01becomes a vertical line, man. So, on
56:02this entire line, x will be the same as
56:04a. It will be a. So that means its
56:06radius is B. Meaning, indirectly, its B
56:09is 10. And if the student is a bit more
56:12educated, it means indirectly its b is
56:15-10. Why take a minus? Because your
56:17circle is basically under the x-axis.
56:19Let me repeat what I said. This point
56:22is a, b, so the radius will always be
56:24at 90 degrees. When it's at 90, on the
56:25x-axis, y is always zero. On a vertical
56:28line, x is always constant. So its x
56:29will also be a. This becomes (a, 0). So
56:32, brother, if this x-coordinate is a,
56:34then the change happened in the
56:36y-coordinates vertically, and obviously
56:38, it went from b to 0. So that means
56:40this gap is 10; that is the length,
56:43though. But location-wise, what should
56:46its sign be? Its sign should be minus.
56:48Meaning this point is under the x-axis.
56:50By the way, under the x-axis. Now tell
56:52me, what is the equation of a circle?
56:55We studied before that the equation of
56:57a circle is this. Look, we call the
57:00equation of a circle (x-a) ²; next, we
57:03say (y-b) ² = radius squared. That's
57:06what we call the circle equation, right
57:07? So you already know this thing: (x-a)
57:09whole squared, copy the same. You know
57:12what your b is. By the way, your b is
57:14-10. From here, I already know what b
57:17is; my b is -10. So when I put -10 in
57:19place of b, it becomes (y + 10) ².
57:23There you go, and the radius, I already
57:24know the radius; this is also 10². So
57:27I found out this new thing that my b
57:29was, my b was -10. So, the coordinates
57:32of the center, the y-coordinate of the
57:34center I already knew was -10. Alright,
57:36now I got stuck here. Suppose I got
57:39confused here. Now we know what x and y
57:43are, right? I told you about x and y in
57:45the introduction. We learned it's any
57:47point on the circumference. What are x
57:49and y? Points on the circumference.
57:52What does x and y show? Points on the
57:54circumference. Points on the
57:56circumference. Because look, what are a
57:57and b? They are the coordinates of the
57:59center. What is R? Radius. So these X
58:01and Y, what are they? These are the
58:04coordinates of, sorry, points on the
58:05circumference. Now, that's why a point
58:07is already given to you. What point is
58:09given? The point given is 5, -16, here
58:11it is. So what will this point do? This
58:13point will satisfy. What? The circle.
58:17This point will satisfy. What will it
58:19satisfy? Obviously, this point will
58:22satisfy the circle. Obviously. Okay.
58:25Now what will you do? You will put this
58:27point into the circle. What will you
58:29put in place of x? 5, there you go.
58:31What will you put in place of y? -16,
58:33there you go. This will become 100.
58:35Here it is. Let's make this even easier
58:37. Apply basic math. (5-a) ² = 64.
58:41Apply the square root on both sides. ±
58:438, let's make it easier. When we
58:46consider +8, what is the answer for a?
58:48-3, and when we consider -8, what is
58:51the answer for a? 13. What is the
58:54answer for a? 13. So from here I found
58:56out that I got double values for a. I
58:58got double values for a. When I got
59:00double values for a, what is the last
59:02step and what does this mean? Your
59:03equation was up to this point. You got
59:05stuck here. Only this thing was unknown
59:07to you. Here it is. (x-a) ² + (y + 10)
59:10² = 100, this was your equation. Now
59:14tell me, there are two possible values
59:16for a. One came out to be -3 and one
59:17came out to be 13. So when you put -3
59:19in place of a, the situation became
59:22something like this. The situation
59:24became something like this. Here you go
59:26. And when you put 13 in place of a,
59:29when you put 13 in place of a, what
59:31will happen? When you put 13 in place
59:34of a, what will finally happen? Here
59:36you go. So when you put 13 in place of
59:38a, it becomes (x-13) ². And (y + 10)
59:43² = 100, this was your target. It was
59:47written as plural, right, equations,
59:50equations. Now look at what this means.
59:52The value of a is double-lined, the
59:54value of a is double-lined. Look, what
59:56was your center? By the way, your
59:58center was (a, b). This was your center
1:00:02, this was your center. Now tell me, if
1:00:06this was your center, this was your
1:00:08center, tell me what is the answer for
1:00:10a?-v, so you already knew b. What was b
1:00:14? b was-so one center can be like this
1:00:16and one center can be like that,
1:00:18meaning these are the two possibilities
1:00:20, these two situations can exist.
1:00:22Meaning your two circles are something
1:00:24like this. Clear? One circle is 13 -10
1:00:28and one circle is -3 -10. Okay? So if
1:00:33if someone wants a proper sketch now, I
1:00:35will show you a proper sketch. This
1:00:37means the situation is something like
1:00:39this. Look, I will show you a slightly
1:00:41more proper sketch now. Okay? So that
1:00:44if any student still has a doubt, they
1:00:46can check it. Okay? They can check it a
1:00:48little. Meaning the situation is
1:00:50something like this. You have you have
1:00:52two two types of circles possible. Two
1:00:55types of two types of circles are
1:00:57possible. I will tell you both of those
1:00:59types. Okay? So now now you got the
1:01:01confirmation, right? Now we are
1:01:02confirmed what exactly happened
1:01:04originally. What all can happen
1:01:06originally. Okay? Now look, there are
1:01:09two types of centers. This means that
1:01:11there are two possibilities. What can
1:01:13it be? One circle is the one whose
1:01:15x-coordinate is positive. A circle with
1:01:18a positive x-coordinate will be
1:01:20something whose center is somewhere
1:01:22here. Look, one circle will be
1:01:24something whose center is positive. It
1:01:26will be something like this. Meaning
1:01:27which circle is this with coordinates
1:01:3013 -10, 13 -10, and you have another
1:01:32circle, one more, and it is also
1:01:35tangent to the circle tangent to the
1:01:37x-axis. It is something like this. But
1:01:40its center, what is its center? Look,
1:01:42meaning this circle here, its its its
1:01:45x-coordinate is a bit larger. Its
1:01:47x-coordinate is a bit larger, so it
1:01:49will be somewhere here. I am just
1:01:50giving an idea, by the way, I am using
1:01:52the word idea repeatedly, and the
1:01:54examiner never said that you should
1:01:55sketch this. Never said that you should
1:01:59sketch this. Clear, kid? Make its size
1:02:02a little bigger. Okay? Meaning the size
1:02:04it has, the same size will be for the
1:02:05other one too. The size, by the way,
1:02:07will be the same for both. This fact is
1:02:09confirmed. Because what is the radius
1:02:10of both? The radius of both is the same
1:02:12. Clear, son? Okay, now look, and the
1:02:15other circle was this one. What was its
1:02:20center? Look, its center is -3, -10, so
1:02:22it could be somewhere here. I'm just
1:02:25giving an idea, by the way, like this.
1:02:27So this means there were these two
1:02:29possibilities. But by the way, what
1:02:31should the size of both circles be? It
1:02:33should be the same. What should the
1:02:35size of both be? The same. So what do I
1:02:37do with this? I'll erase it and copy
1:02:39the same size. Because what is the
1:02:41radius of both? The radius of both is
1:02:43the same. The radius of both circles is
1:02:45the same. Okay? So if the radius of
1:02:47both is the same, then what should the
1:02:49size be? It should be the same. Nowhere
1:02:51in the question does it say that you
1:02:53have to sketch it. It doesn't say that
1:02:55anywhere. Okay? It is not written
1:02:57anywhere. There could be such a
1:02:59situation. It doesn't say anywhere.
1:03:01Alright? I'm just showing extra work to
1:03:03be on the safe side. Extra, clear son?
1:03:07Some situation like this, some
1:03:11situation like that. Clear, son? So,
1:03:14there can be two possibilities; your
1:03:17circle could be this one or this one.
1:03:20Now, where is 5, -16? It could be this,
1:03:23it could be that. Clear? 5, -16 should
1:03:28be right here, it should be below,
1:03:30right? So, 5, -16 should be this point.
1:03:34Clear? Clear, son? Let this be 5, -16.
1:03:36There you go. I have given you an idea.
1:03:39Done. So, in the next class, God
1:03:41willing, we will try some slightly more
1:03:43complicated questions. Alright? The
1:03:46questions we did today also included
1:03:47basics. Some were also very easy. But
1:03:50the questions we try in the next class
1:03:51will be a bit tougher. Alright, son?
1:03:54Goodbye.

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