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AS Level Mathematics 9709 | Coordinate Geometry 9 | Standard Equation of Circle | Raja Shafqat — Transcript

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0:08Assalamu Alaikum, beta. In the last
0:10class, we were studying the standard
0:12equation of a circle. In fact, we did
0:14some questions from past papers
0:15regarding that. And we did some
0:17practice questions. Let's do a little
0:19more practice today. Then, Insha'Allah,
0:21we will complete all the points one by
0:23one. Then finally, we will practice the
0:25overall past papers. Alright, beta? We
0:27will practice the overall past papers.
0:28Come on, now finally look at what the
0:30statement of the first question is. The
0:32examiner says the equation of the
0:34circle is already given as this. Show
0:37that point A lies on the circle. First
0:41point, number two, find the equation of
0:43the tangent to the circle at point A.
0:45Equation of the tangent. Two things.
0:48Let's start slowly, seeing what hint
0:50the examiner is giving us first. Right?
0:53What hint is the examiner giving us
0:55first? We will read it one by one. The
0:58very first thing, what is the equation
0:59of the circle that you and I have
1:01studied so far? Right? What is the
1:03equation of the circle that you and I
1:04have studied? Look, first we studied
1:07that we read this equation. Standard
1:09equation of a circle. x-a² + y-b² =
1:17radius squared. This is the equation of
1:20the circle we studied. What do we call
1:22this equation? We call this equation
1:24the standard equation of a circle. We
1:27always call this equation the standard
1:31equation of the circle. Right?
1:35Previously, you and I have only studied
1:36this so far. Standard equation of a
1:39circle. Now, look at the equation you
1:41have been given. x-3² is given. And
1:46what is given next? Next is given y + 2
1:50squared, y + 2 squared = 25 = 25, so
1:54first compare this with that. From here
1:58, we found out what a is. After
2:00comparing it with the standard form, I
2:03found out what its a is. Its a is 3.
2:06Clear, beta? We compared it. From here,
2:09I found out that its-b is 2. Its-b is 2
2:12. So, that means what is its b? Its b
2:14is -2, and what is the radius squared?
2:19The radius squared is 25, so it means
2:21when the radius squared is 25, then
2:22what is the radius? The radius is five.
2:26So, first we only did this work, and by
2:28comparing, I found out that this is the
2:30center of my circle. What is the center
2:32of the circle? 3, -2 is the center of
2:34my circle. And what is the radius of my
2:37circle? Its radius is five. I have a
2:39rough idea now of what the examiner is
2:42trying to do. Is that okay, son? I have
2:44a rough idea of what the examiner is
2:45trying to do. Clear? Let's continue and
2:48see what our task is. Alright, son.
2:51What else is our task? Come on, we have
2:53found the center and the radius. What
2:56do we do next? Finally, I will just
2:59show you a rough sketch. Suppose this
3:01is a circle. Okay? This is its center.
3:04I just plotted a rough sketch like this
3:06. I marked the center of the circle.
3:08This is not a proper circle. I am just
3:10giving a random example. So, the center
3:13of this circle is 3, -2. Here it is, 3,
3:17-2, that is the center of the circle.
3:21Okay, what is the radius? The radius is
3:24five. Done. Now let's start reading the
3:26question. Number one. Show that this
3:28point lies on the circle. This point
3:31lies on the circle. Okay? And one more
3:35thing, then find the equation of the
3:37tangent to the circle at this point.
3:39The same point that we have to show is
3:41on the circle. We have to draw a
3:42tangent line at that point. Meaning, if
3:44this is three, then 4, 5, 6 will be
3:45somewhere here. Six will obviously be
3:47further ahead. And what is its
3:48y-coordinate? y is -6, meaning it will
3:50be a bit lower. So I just marked a
3:51point like that. Here you go. I just
3:54randomly marked a point here. Okay, son
3:57? So I always use the word that I am
3:59showing a rough sketch. So suppose I
4:02have to show that this point is on the
4:04circle. My target is to show that this
4:06point is on the circle. This is my
4:08first target. This is my target. Okay?
4:11Show the point on the circle. So it is
4:13very easy. Now tell me, what equation
4:15of the circle is given? This is given:
4:17the equation of the circle (x-3) ² + (
4:20y + 2) ² = 25, here it is. = 25, is
4:23that clear, son? Now tell me, we have
4:27already studied that any point on the
4:29circle does what to the circle?
4:32Satisfies it. We have studied in math
4:36that any point on the circumference
4:39will do what? It will satisfy the
4:43circle. Obviously, the point that is on
4:45the circle will do what? It will
4:47satisfy the circle. It will do what? It
4:51will satisfy the circle. So, when we
4:53study coordinate geometry in math,
4:55satisfying an equation means you plug
4:57in that point. The left and right sides
4:59will balance out. The left and right
5:01sides will balance out. Another way is
5:03that if you find the distance between
5:05these two points, what will that
5:07distance equal? It will equal the
5:09radius. Okay? So the first thing you
5:11should do is, one option is to plug in
5:13this point. 6, -6 in the question. As
5:15soon as we plugged in 6, -6, what was
5:17the answer to the question? 3²-4²,
5:19when simplified, it turned out that 25
5:22= 25, meaning the left and right sides
5:25were what? They balanced out. It is
5:27proven that the point is on the circle.
5:29That's it. It is proven the point is on
5:31the circle. The point is on the circle.
5:34This thing is proven. This is what we
5:35had to prove, by the way. Clear, kid?
5:38And this is exactly what we had to
5:40prove, that the point is on the circle.
5:42Done, kid? Let's move on, half our work
5:44is done. Next. The next task. What's
5:48next? Find the equation of the tangent
5:49to the circle at the point. Equation of
5:50the tangent. I am showing you a rough
5:53sketch. A rough sketch. I am just
5:55randomly drawing a line like this. Look
5:57at this line, what is your target? This
5:59is what needs to be shown. Your target
6:00is to find the equation of the tangent.
6:02So, this line that I just drew, this
6:05line touches the circle at point A.
6:07This line touches the circle at point
6:09A. What is your target? This line is
6:12the tangent. So I will mark here that
6:15this line is the tangent at A. This
6:17line is the tangent to the curve, the
6:20tangent to the circle at point A, the
6:23tangent to the circle at point A. We
6:25studied a property that the radius is
6:28always perpendicular to the tangent
6:31line. When I join the center with the
6:34point of tangency. When I join the
6:36center with the point of tangency, what
6:38will always be formed? 90 ° will
6:40always be formed. Clear, kid? So we
6:42applied a property. What is the
6:44property? That the radius is always,
6:46the radius is always perpendicular to
6:49the tangent line. Right? We studied a
6:52property that the radius is always
6:55perpendicular to the tangent line. So
6:58we will have to apply this property.
7:01The radius is always perpendicular to
7:03the tangent line. Meaning we applied
7:06that property, the radius-tangent one.
7:08We applied the radius-tangent property.
7:11Now tell me one thing. The very first
7:13thing, if I ask you what the gradient
7:15of AC would be? So, what is the formula
7:17for gradients? The gradient formula is
7:20y2-y1 over x2-x1, so I applied the
7:22gradient. So, it becomes y2-y1 over x2-
7:26x1, which is -4/6-3; that is the
7:30gradient. That is the gradient. Whose
7:32gradient is this? It is the gradient of
7:35AC. We got the gradient of AC. Okay? So
7:37, if I ask you what the gradient of the
7:39tangent line will be? This is the
7:41gradient of AC. So, my target is to
7:43find out the gradient of the tangent
7:45line. What does the radius always form?
7:48It forms 90 ° with the tangent line.
7:51So that means if you have its gradient
7:52as -4/3, then what will be the gradient
7:54of the tangent? 3/4, the negative
7:57reciprocal. What should its gradient
7:59have been? The negative reciprocal. Now
8:02, what is your target? Your target is
8:04the equation of the tangent at. Your
8:07target is the 'equation of the tangent
8:11line at A'. This is your target, the '
8:15equation of the tangent line at A'.
8:18This is your target, by the way. Okay,
8:20now tell me one thing, for a line, you
8:22require a point. You already have point
8:24A, here it is (6, -6). You have the
8:27point already, and the gradient. You
8:28already have the gradient. What is the
8:30gradient? 3? 4, here it is. Now, we
8:33have to use coordinate geometry here:
8:36y-y 1 = m times (x-x 1), here it is.
8:39Make it a little easier, so what will
8:41this become? This became 3/4x, minus
8:4518/4, and minus 6. Done, son? Now, we
8:49just need to simplify this a little bit
8:50. Okay? 18/4 means 9/2, which is -4.5,
8:54so what is the answer? 3/4x-10.5, or
8:5821/2; this is the final answer derived
9:02for your question. So what was your
9:05target? By the way? Your target was the
9:08equation of the tangent line at point
9:10A. Done. So here we used the condition
9:12that what is the radius always? The
9:14radius is always perpendicular to the
9:16tangent line. The radius is always
9:18perpendicular to the tangent line.
9:19There is no restriction. So you can
9:21show the line equation in any form.
9:24There is no restriction at all. By the
9:25way. There is no restriction of any
9:27kind. Is it clear, son? So the
9:29condition we used here was the radius
9:31and tangent line condition. Done, son,
9:34the radius and tangent line condition
9:36was used. Next, let's move to another
9:38question; please read the question well
9:40first to see what the examiner is
9:41asking. Let's say the line cuts the
9:44x-axis at A and the y-axis at B, let's
9:47just stop right here. Let's stop here
9:49for a minute. So look, a line is given
9:51to me by the way. Okay? Slowly, slowly
9:54read it all: 2x + 5y = 20, this is your
9:58question. If you want, you can
10:00rearrange it a little. Of course, if
10:02you want to arrange it, otherwise it
10:03will be solved even if you don't
10:04arrange it. Here it is. If I
10:06deliberately rearrange this a bit, it
10:09becomes -2/5x + 4, there it is. I
10:12intentionally rearranged it. It is not
10:14written anywhere in the exam that you
10:16must rearrange it first. It's not
10:18mentioned at all. So I did it so that I
10:20could show you that the line is
10:21basically of this type. It is a line of
10:24this type. By the way. Meaning, this is
10:27the situation. This, I deliberately
10:29sketched and showed it to you. It is
10:31not written anywhere that you have to
10:32sketch it. Now look at what is written
10:34in the question. Do you know what the
10:36y-intercept of this line already is?
10:37You already know the y-intercept of
10:40this line. Okay? You already know the
10:43y-intercept of the line without working
10:45. It is 0, 4. The y-axis is at B. So by
10:48the way, what is this B? B is 0, 4,
10:50there it is. Here it is. Let's do it
10:53like this, that on the y-axis, x is
10:55always zero. So when you set the value
10:56of x to zero, what will y automatically
10:58be found to be? Y will automatically be
10:59found to be 4. What will y
11:00automatically be found to be?
11:01Automatically, y will be found to be 4.
11:03This B is found, simple. Well, one more
11:07thing, the line is cutting the x-axis
11:09at A, the line is cutting the x-axis at
11:11A, the line is cutting the x-axis at
11:13point A, so now I have to find point A.
11:16This is your point A. Here, I marked it
11:18. I highlighted it. This is point A.
11:20Now I have to do the work for A, so
11:23tell me one thing. On the x-axis, on
11:26the x-axis, what is y? Tell me. y is
11:29zero. So if I put zero in place of y,
11:31what will x be? x will be 10. Clear,
11:34kid? So that means what is your x? Your
11:36x is 10. So that means what are the
11:38coordinates of A? The coordinates of A
11:40are 10, 0, giving you a graph idea, by
11:42the way. Giving you such a graph idea,
11:44so what did I do first? I have found
11:46out what b is and what a is. I have
11:49found out what my b is and what my a is
11:51. Now let's read on to see what our
11:53main target is. Alright, we have found
11:56out what b is. And what a is. Okay?
11:58Just for reference, I only showed it to
12:00you; the line wasn't required. By the
12:02way, the line cuts the X-axis at A (
12:04done), X-axis at A, Y-axis at B (done),
12:07Y-axis at B, and point C is the
12:09midpoint of line AB. Enough, let's stop
12:12here and look at this part: where is
12:14point C? Point C is the midpoint, here
12:16it is. The examiner is saying that its
12:18midpoint, who is that midpoint? That
12:20midpoint is C, by the way. So now, if I
12:22ask you how to get the coordinates of C
12:24? C. So what is C? A hint is given in
12:27the question that C is basically the
12:29midpoint of line AB. Here is the
12:32midpoint. So how will you find the
12:34coordinates of midpoint C? What is the
12:36midpoint formula? The midpoint formula
12:39is (X1 + X2)/2, (Y1 + Y2)/2, so how
12:41will the midpoint be calculated? X1 +
12:44X2/2 and similarly Y1 + Y2/2, so what
12:47coordinates for C were found from here?
12:51The coordinates of C were found to be 5
12:54and 2; these are your coordinates of C.
12:56These are your coordinates of C. There
12:59you go, C is found. It's not solved yet
13:01. Slowly, slowly, we are solving it.
13:03Now come and read. Find the equation of
13:05the circle that has center C and that
13:07passes through A and B. Okay? Now, is
13:11the situation a bit clearer as to what
13:13the examiner wanted to do? Okay? Now I
13:16will show you this thing again. Now I
13:18understand what the examiner wanted to
13:19say. Look at this. I will show you this
13:21same thing here. So, what was the
13:23examiner's message? Now he is showing
13:25it to me like this. He says to tell the
13:27equation of the circle that has center
13:29C and passes through A and B, which
13:32means the situation was something like
13:34this. Look, the situation was like this
13:37, that your circle is passing through
13:40where? Your circle is passing through A
13:42and B. There you go. Let's make it a
13:43bit easier. There you go. This is your
13:45circle, I've made the situation a bit
13:47more obvious now. Here you go. Now I've
13:49drawn the circle. The situation is a
13:51bit clearer now. The situation is like
13:54this. Done, son. Now, why did I show
13:56this sketch? So that it becomes a
13:58little more obvious. Okay? Look again.
14:00Let's look again, son. These were your
14:03coordinates for B. This was the
14:04coordinate B. Look, I just copied that
14:06same circle here. Okay? Let me plot
14:08this a bit more properly. Here you go.
14:11The situation is something like this.
14:13This is the situation. Done. Okay. You
14:15already know the coordinates of A. What
14:17were the coordinates of A? By the way.
14:19Look, what were the coordinates of A?
14:20This was your point A. Point on the
14:22x-axis. The coordinates of A were 10, 0
14:24. The coordinates of A were 10, 0,
14:27there you go. The coordinates of A were
14:2910, 0, there you go. And what were the
14:31coordinates of B? The coordinates of B
14:33were 0, 4, here it is. The coordinates
14:36of B were 0, 4, I have already found C.
14:38What is C? By the way, the coordinates
14:40of C are 5, 2. Here you go. I already
14:42know the center as well. This was the
14:43center of the circle. What is the
14:45center of this circle? 5, 2, here you
14:47go. So, I have this information. Now
14:50let's read the question. What needs to
14:51be done? First, find the equation of
14:54the circle. Okay? Find the equation of
14:56the circle. Tell me one thing. What is
14:58the equation of a circle? By the way,
14:59we have talked about this many times.
15:01The equation of a circle is (x-a) ² +
15:04(y-b) ² = radius squared. This is the
15:07equation of the circle. This is the
15:09equation of the circle, by the way.
15:10Clear, son? This is the equation of the
15:12circle. Okay. Now I already know the
15:14center of the circle. The center of the
15:16circle is A and B, so who is A? A is 5,
15:18and who is B? B is your 2. Here you go.
15:21I know both A and B. Done. What do I
15:23not know? I don't know the radius, what
15:25is the circle's radius? So the radius
15:27can be found easily. If you need the
15:29circle's radius, radius means distance.
15:32So how will the radius come? Whatever
15:34the distance is between B and C, that
15:35will be the distance between C and A.
15:37And both these distances will
15:38indirectly show the radius. Because the
15:40gap from the center to the
15:42circumference is what we call the
15:44radius. We call that gap the radius. So
15:47what will the radius be? x2-x1 squared,
15:49y2-y1 squared, so what is the answer
15:52for the radius here? The radius answer
15:55is 29. There is nothing very logical
15:57about it. Here you go, the radius
15:59answer is 29. So what will be the
16:01square of 29? 29 What will be the
16:04square of 29? 29, done. Now read
16:06further. Well done. And what else is
16:08written? Our target is achieved. This
16:10is the equation of the circle. The next
16:12equation of the circle has arrived.
16:13Next, show that the circle also passes
16:16through the origin. The circle also
16:18passes through the origin. There are
16:20two options. This means if the circle
16:22is passing through the origin, then
16:23what will the origin do? It will
16:25satisfy the circle. The second thing
16:27could be that if you have to show it,
16:29you have another option as well. It is
16:31saying, what is it saying? Show that,
16:33show that the origin lies on the circle
16:36. The origin lies on the circle. It is
16:38saying to show it. So it means that if
16:40the circle is passing through this
16:42point, then the distance between the
16:44origin and the center, the distance
16:46between the origin and the center
16:48should be what? It should be the radius
16:50. So you do one thing. You do one thing
16:53. Now what will we do? Let's find the
16:54distance between O and C. So if I apply
16:56the distance formula, the distance of O
16:58and C, what will it be? 5-0 squared.
17:025-0 squared, 2-0 squared, so indirectly
17:05what did its answer derive to?
17:08Indirectly, its answer also derived to
17:09this. Same 29, this was to be shown
17:12that if this point origin was on the
17:15circle, then it would satisfy the
17:17circle. The gap from the center to the
17:19origin would be equal to what? That gap
17:22would be equal to the radius. That gap
17:24would be equal to the radius. Done, son
17:26? So this was also an easy question.
17:28Whatever was required in this, we
17:30proved one by one. After a while, I
17:31will get two or three tough questions
17:33done. Okay? Come on, let's try the next
17:35question. Just read the wording of the
17:37question. I am waiting for you. Come on
17:39, read the language correctly, son,
17:41what the examiner is asking us. Okay?
17:43Read the language first. A circle
17:45passes through this and this point.
17:47Wait. Did you note one thing? Both have
17:50the same y-coordinate. Meaning if 3, 2
17:52is somewhere here, then 7, 2 is ahead
17:54of it. Because, because both of their
17:56y-values are the same. And has a radius
17:59of this. Its radius is this 2, 2, find
18:01the two possible equations of the
18:03circle. It is restricting that two
18:05equations can be possible. So first,
18:07let me show you a rough sketch. What
18:09does the examiner want to do with the
18:10rough sketch? Let's slowly start moving
18:13forward. What message did the examiner
18:15give you? The examiner gave you this
18:17message. The examiner gave you this
18:19message. Here is one point. Here is one
18:21point. Okay? Come on, let's look at
18:22this first. Watch carefully. Tell me,
18:24one of your points is 7 2, here it is.
18:277 2, this 7 2, and you have one more
18:30point as well. You have one more point
18:34too. Alright, 7 2 is here. Here is 7 2,
18:37so what was the first point given? That
18:40was 3 2, so first of all, by the way,
18:42this is a horizontal line. Okay? Here
18:46is 3 2, I have marked this point on the
18:48diagram as well. I showed these
18:49sketches so you gain some confidence.
18:51What is the center? Don't know. What is
18:53the center? Don't know. I don't know,
18:55it must be something. I don't know.
18:57Obviously, it must be something. What
18:59is the given radius? By the way, if you
19:00have read the question, what is the
19:02radius given? The radius is given as 2
19:042. Here, its radius is given as 2 2.
19:07What is its radius given as? 2 2 radius
19:10is already given. The radius is already
19:13given. Now, what is the examiner asking
19:15you? Tell me one thing. When I drop a
19:18line from the center here, we know a
19:20property we studied that a line through
19:22the center bisects a chord. We studied
19:25a property that a line through the
19:26center bisects the chord. So, I am
19:28mentioning for reference here that I
19:31used a circle property. We studied a
19:33circle property that a line through the
19:35center bisects the chord. We studied
19:38the property of a line through the
19:40center. A line through the center
19:44bisects the chord. A line through the
19:48center bisects the chord. And anyway,
19:50if you think that this triangle is an
19:52isosceles triangle. It is an isosceles
19:54triangle. So if the triangle is
19:56isosceles, then anyway, when you drop a
19:58line, it bisects the base. Now, tell me
20:00one thing, when I dropped this
20:02isosceles line, I know that on this
20:04whole line, the value of y is fixed.
20:06The value of y is two. So it means that
20:09if I tell you that this point here,
20:11this point. Come on, look at this point
20:13, here it is. I will highlight it. This
20:16point, the point I am marking, I don't
20:18know what the coordinates of this point
20:20will be, let's suppose, okay. Now tell
20:23me one thing, obviously if I divide
20:24this into equal parts, then I will get
20:26its midpoint. Let's name this point M,
20:29so how will M be calculated? Tell me, M
20:31is the midpoint of what? M is the
20:33midpoint, the midpoint. Let's name this
20:36point, suppose, you name this point A,
20:38name this point B, okay here, so this
20:40is the midpoint of AB. By the way, let
20:43me find out the midpoint from here. How
20:45will the mid, the midpoint come? The
20:46midpoint will be x1 + x2/2 and anyway,
20:49there is no need to find the
20:51y-coordinate. Well, I will do it anyway
20:53because the y-coordinate will be the
20:54same for everyone. So what is the
20:56answer to this? The answer is 5, 2,
20:58here you go 5, 2, so that means the
21:00coordinates of the midpoint are 5, 2,
21:02here it is. This is 5, 2, let's go. So
21:06now we know one thing that on this
21:07entire vertical line, this whole line
21:09is vertical, right? The line is
21:10perpendicular to the chord. So that
21:12means if I ask you what the coordinates
21:14of the center will be? What will be the
21:16coordinates of the center? Then what
21:18will you say? It is confirmed that the
21:20x-coordinate of the center will be five
21:21. What is the y-coordinate? I don't
21:23know that. What is the y-coordinate? I
21:25don't know that. But it's obvious, why
21:27will x be five? Because on a vertical
21:29line, x is always constant. What is the
21:32equation of a vertical line anyway? x =
21:34constant. Now look at what your
21:36question statement is. Your question
21:38statement is. Find the two possible
21:40equations of the circle. Okay, now tell
21:42me one thing, I don't know what the
21:44equation of the circle is. I know that
21:45this gap is also the radius. This gap
21:47is also the radius. So AC is also the
21:49radius and BC is also the radius. Isn't
21:51it like that? AC is also the radius and
21:53this is also the radius. So you do one
21:55thing, tell me what the gap between A
21:58and C will be? The length of this A and
22:00C, what will that length obviously show
22:02? It will show the radius. The length
22:04will obviously show the radius. So do
22:05one thing, first apply the distance
22:08formula on AC. What will the distance
22:09formula be? x2-x 1², 5-3², 2² + y2-y
22:131², b-2² is equal to the radius. And
22:18what is the radius? The radius you have
22:20is 2. You have the radius, it is 2√2.
22:25Now what should you do? Square it.
22:27There you go, squaring both sides, so
22:29this will be 4 and this side will be (b
22:32-2) squared; when you square this, it
22:34becomes 4*2, which is 8. Its answer is
22:38eight. Make it even easier, this
22:40becomes (b-2) whole squared equals 4;
22:42taking the square root on the right
22:44side, it becomes ± 2, so two answers
22:46for b are derived from here; one value
22:49of b is zero and one value of b is 4,
22:51which means the center's coordinates
22:53can be of two types. So as soon as I
22:56got B, it immediately clicked what the
22:58circle's center was. What was the
23:01center of the circle, 5, B? 5, B, so if
23:04B is zero, then the circle's center
23:06will be 5, 0; the center of the circle
23:09will be 5, 0, is that clear, kid? And
23:13similarly, if your B, if your B is 4,
23:15then the center of the circle will be (
23:175, 4). Here you go, the center of the
23:19circle will be (5, 4). These are the
23:22two possibilities you have for the
23:24center. Look, so two types of circles
23:27came up; one circle whose center is (5,
23:300), okay? And there is another circle
23:32whose center is (5, 4). What is the
23:34radius? I already know the radius for
23:37both, the radius is 2√2, 2√2, I
23:39already know the radius, the radius for
23:42both circles is 2√2. Here it is. What
23:44were you asked? Two possible equations
23:46of the circle. So we know what the
23:47equation of a circle is, right? x-a²
23:51y-b² = radius squared. So now put both
23:56centers in one by one. So when you put
23:595, 0, a becomes 5. b becomes 0 and the
24:02radius squared becomes 8. This is one
24:05answer. This is one answer. And there
24:08is one more answer. When you put 5, 4,
24:11it becomes x-5 squared + y-4 squared =
24:14radius squared, equals radius squared,
24:17what is the radius squared? Eight, here
24:21it is. The radius squared is eight.
24:23What do you have to do with these two?
24:24This was your basic target. What did
24:26you need? You needed two possible
24:29equations of the circle. Two possible
24:33equations came because there are two
24:35possible options for the center, the
24:37center can be 5, 0 and the center can
24:40also be 5, 4. That is why the examiner
24:42used the word two possible equations
24:44for the circle. Done, son, for this
24:46circle. Done. So we tried another small
24:49question. Let's try more questions.
24:52Read the question. What does it say?
24:54Read it carefully. A circle passes
24:56through this, this, and this point.
24:58Show that OA is a diameter. OA is a
25:01diameter. Okay? First, we have to read
25:03the question to see what the examiner
25:05wants to do. Number one, look, what is
25:08his main message? OA is a diameter.
25:12Okay? He is just saying the circle
25:13passes through this, this, and this
25:15point. I am just giving you a rough
25:16idea. Somewhere here will be 0, 1, 2, 3
25:18, 4, 5, 6, somewhere here will be 6,
25:20and somewhere here will be 8, 4. The
25:22circle passes through all of them like
25:24this. The circle passes through all of
25:26them like this. Clear, son? I am just
25:29giving you an idea. By the way, an idea
25:31. And the examiner says this is your O.
25:33This is your A and this is your B. Wait
25:34, no, the circle passes through these
25:36three points. The circle passes through
25:37this, this, and this. Show that OA is a
25:40diameter. Okay? Okay? Show that OA is a
25:44diameter. Okay? So first I have to show
25:47what OA is. OA is a diameter. Okay? OA
25:51is a diameter. Okay? Let me check what
25:53condition the examiner wants to apply.
25:56What condition does he want to apply? I
25:58will guide you. Okay, son? Let's go.
25:59Now look, number one, OA is a diameter.
26:02I just plotted a line through the
26:05center for an idea. There you go. So it
26:08means what is this? The center of this
26:10circle. Let me mark it here. What is
26:12the center? Don't know. I don't know
26:14yet. The examiner says show that OA is
26:16a diameter. There it is. OA is a
26:17diameter. So what is this? Eight, what
26:19is it? Zero. 8, 4 A and the origin is
26:21what? 0 0, okay? Good. First show that
26:26this is a diameter. This is what the
26:28examiner is saying. He says the circle
26:31passes through this, this, and this.
26:33Show that OA is a diameter. Okay? Let's
26:36confirm right now. There is another
26:37point on the circle, too. There is one
26:40more point. That point you have is 6,
26:42so it must be somewhere around here too
26:44. It would be before eight, right?
26:46Obviously, if this is zero and this is
26:47eight, then six will be somewhere in
26:48between. So, suppose I mark another
26:50point here. Here is the point. Suppose
26:53I marked this point. Okay? Let's just
26:55mark another point randomly like this.
26:57Okay? So, I am showing this for
26:59explanation. Here you go. So that means
27:00this point here, which point is this?
27:02This point is your 6, right here. This
27:05point is 6. Here you go. Now, the
27:08examiner says to show that OA is a
27:09diameter. So then, you must know a
27:11property of circles. If this is a
27:13diameter, it means when both points of
27:15the diameter are joined at B, what
27:17angle should be formed? 90 should be
27:20formed. Look at it again. If this is,
27:24if this is a diameter, so if OA is a
27:26diameter, then when both points of OA
27:28are joined, when both points of OA are
27:31joined with B, what should be formed?
27:3490 ° should be formed. Okay? Meaning,
27:37which property is applied here? The
27:39semi-circle property is applied here.
27:41The examiner is indirectly making you
27:43apply the semi-circle property.
27:45Indirectly, the examiner is making you
27:48apply the semi-circle property. What
27:51was the semi-circle property? The
27:53semi-circle property was that whenever
27:55you have a semi-circle, when you join
27:57both the end points of the diameter, it
28:00forms 90. So, we used to read this
28:03property as the angle subtended by the
28:05diameter, the angle subtended by the
28:08diameter, is always complementary.
28:13Let's say is always 90 °. Forget
28:15complementary, it's 90 °. We had
28:17studied this as a property of a
28:19semi-circle. Property of a semi-circle.
28:22If OA is a diameter, then easily we
28:24have to prove angle B. Sorry, we have
28:26to prove point B, that the angle there
28:28is 90. So, take help from coordinate
28:30geometry. Now, find out the gradient of
28:33OB; 6/6 comes to 1. Meaning Y2-Y1 over
28:36X2-X1, so we found out that its
28:38gradient is what? It came to one. The
28:40gradient of OB came to 1. Similarly,
28:42now look at the gradient of AB. So,
28:44what is the gradient of AB? The
28:46gradient of AB is Y2-Y1 over X2-X1,
28:49here it is. So, this will come to -1,
28:51here it is. So, from here, we found out
28:53that the product of both these
28:54gradients is what? -1. The gradient of
28:56OB and the gradient of AB is what? AB
28:59is -1. So, that means that angle OBA is
29:04a right angle. Angle OBA is a right
29:07angle. This is what had to be shown.
29:10This is what had to be shown, right? If
29:11this is proven to be a right angle, it
29:14means that OA is confirmed to be the
29:16diameter. OA is confirmed to be the
29:18diameter. If this is proven. Next, find
29:20the equation of this circle. Now it's
29:23easy. Now, the equation of this circle.
29:25So, now tell me, what is your target?
29:27Now your target is the equation of the
29:29circle. Now your target is the equation
29:31of the circle. So we know what the
29:33equation of the circle is, right? By
29:34the way, this is an easy question. What
29:36is the equation of the circle? The
29:38equation of the circle says x-a²,
29:41y-b² = radius squared. This is the
29:45equation of the circle. Now, how do we
29:47get the center of this circle? The
29:49center will be the midpoint of the
29:51diameter, i.e., OA. Midpoint of OA,
29:54here is the midpoint of OA. So when you
29:57find the midpoint, X1 + X2/2, here it
29:59is. Next, Y1 + Y2/2, here it is. Now
30:03add these two and divide by two. So
30:05from this, I found out what the
30:07coordinates of the midpoint are. 4 2,
30:09here it is. You have the coordinates of
30:11the midpoint now. Okay? So later, I
30:13found the coordinates of the center.
30:15Here you go, 4 2, it's right here. This
30:16is 4 2, now only one thing is missing,
30:19I don't know its radius. OC is also a
30:22radius. CA is also a radius. CB is also
30:24a radius. These are all radii. The gap
30:27from the center to the circumference is
30:29called the radius. So you need the
30:31radius. Then you do one thing. Do any
30:34one thing. Okay, kid? Just one thing.
30:36Okay? So what did we do? Suppose I find
30:40out the distance OC. Here you go.
30:43Distance between O and C, okay? So, O
30:46is easy for calculations, that's why I
30:49found OC. OC, so 4-0² and 2-0², here
30:54it is. So what was the answer found?
30:57The answer was found. 20, the answer
31:00was found to be 20. 20. This is found.
31:08What is your target? Your target is the
31:10equation of the circle. Now the entire
31:12target is achieved. This is the
31:14equation of the circle. x-a² + y-b² =
31:17radius squared. Done, kid. Now tell me,
31:21what is a? What are a and b?
31:23Coordinates of the center. So I know
31:24what a is. It's 4, and what is b? Your
31:28b is two. Because the coordinates of
31:30the center are called a, b. And what is
31:31the radius squared? The radius squared
31:33is 20. You have achieved your target.
31:36What was your target? You had to find
31:38the equation of the circle. So you had
31:41to find the equation of the circle. Is
31:43that clear, son? That was our final
31:45target for this question. Let's try
31:48some more. Yes, this one is a bit tough
31:50. Okay? Now this is a slightly tougher
31:52question. Go ahead, you read it. Now I
31:55will guide you through it. The same
31:57type of question has appeared in past
31:58papers as well. The same style. The
32:01same style. Clear? A circle passes
32:03through this, this, and this point.
32:04Find the equation of the circle. This
32:06is a very interesting and different
32:08question. Okay? Right? Okay? Now, you
32:10are given a circle. Right? You are
32:12given a circle. Here you go. This
32:14circle is passing through three points.
32:17One point is O. I will just mark a
32:19point like this. Suppose there is a
32:20point O somewhere here. I'll just mark
32:22a point randomly. This O 3, 9 will be
32:24somewhere further ahead. Obviously,
32:27suppose it's somewhere here. And 11
32:29will be somewhere even higher than this
32:31. You can just mark these two-three
32:32points anywhere, man. No issue. It's a
32:34rough sketch, by the way. Suppose I
32:36marked one point somewhere here. Here
32:38you go, one point. Suppose I marked it
32:40somewhere here, just as a random
32:42example. There it is, I marked this for
32:44a rough idea. Okay? Just a rough idea,
32:47I marked a point like this, here you go
32:48. Okay? Mark it slightly further ahead
32:52from this or anywhere else. We don't
32:54need accuracy, right, son? I'm not even
32:57plotting an accurate diagram. And a
32:59third point is also given, what is that
33:00third point? 11, 11 Okay? A bit higher
33:06point, suppose this 11, come on, these
33:07are all rough sketches. By the way. So
33:10I marked these three roughly as well.
33:13The circle is passing through these
33:14three points, by the way. Okay? This
33:16condition is already given in the exam.
33:18Here you go. Oh, one point is named the
33:20origin. O, this point is 0. Here you go
33:23. The coordinates of this point are 0.
33:25Here you go. The coordinates of this
33:27point are 0. Here you go. What was the
33:30next point? One more point was P. 3, 9,
33:35there it is. Who is the third point? Q.
33:38Who is Q? 11, 11 This is the situation.
33:45There is no hint given here. Now you
33:47have to use your own common sense. This
33:48is a very interesting and different
33:50question. The same question appeared in
33:51a past paper for six marks. The same
33:53question has come in a past paper for
33:55six marks. Now look at one thing. We
33:57studied a property. I will repeat it
33:59again. The property was this. A line
34:02through the center bisects the chord.
34:05Is that okay, son? We studied a
34:07property. I will have to use that
34:09property. Okay? Here, let me show it
34:11again. Here it is, I will mark the
34:13property here that we studied. This was
34:15the property we studied. A line through
34:17the center. A line through the center
34:22bisects the chord. A line through the
34:25center bisects the chord. Okay? We
34:27studied this one property. We will have
34:29to use this property here just for
34:31reference. Now I have. Okay, what is
34:33our target? We need the center. The
34:35issue is this. The issue is that there
34:38is no center of the circle. Don't know
34:39what it is. If the center comes, the
34:41radius will come, right? All these gaps
34:43will be the radius. There is no center
34:44at all. So what is the first thing we
34:46will have to do? If I join these two
34:48points, please listen, son. If I join
34:51these two points, where will their
34:53perpendicular bisector pass through? It
34:56will pass through the center. It's
34:58obvious. Whatever, suppose this is the
35:00center of the circle. Suppose it has to
35:01be somewhere, right? I have marked a
35:03center of the circle here. Look, see it
35:04again. Okay? Look again. Here it is.
35:07Suppose your center of the circle is
35:08somewhere around here. It has to be
35:09somewhere. I marked a point. Here you
35:11go. Suppose this is the center of the
35:13circle. Suppose, so when you drop a
35:15line from the center, what will that
35:17line do to the chord? Bisect. So I will
35:19write it here just for reference. Who
35:21is this line? This line is basically
35:25the perpendicular bisector of the chord
35:28PQ. Let's say this is the perpendicular
35:31bisector. We will need one more
35:33perpendicular bisector too. Any other
35:35one, let me plot one more. Here, when I
35:38draw the perpendicular bisector of
35:40these two chords. Here it is, so their
35:42perpendicular bisector will also pass
35:43through where? Look, if I join this,
35:47what? I joined O and P. Here, I joined
35:50O and P, so their perpendicular
35:52bisector will also pass through where?
35:55It will pass through the center. That's
35:56how it is, isn't it, son? Where will it
35:58pass through? It will pass through the
36:00center. So what did I do? I plotted two
36:03chords. Okay? Let me explain this again
36:06. I've drawn another line like this.
36:08Okay? I am just giving you a rough idea
36:10. Another line like this; by the way,
36:11how should this line be? This line
36:13should be slightly tilted. Okay? The
36:17line should be tilted a little. Like
36:19this. Just a rough sketch I've shown.
36:23So, I will bisect this one as well.
36:25Like this, here it is. I will bisect
36:27this one too. Right, here it is. Done,
36:29son. So, I will mark it on this. What
36:31is this? This is the perpendicular
36:35bisector. Perpendicular bisector of the
36:38chord OP. Of the chord OP. You will
36:42have to apply this property. Because
36:44you know that its perpendicular
36:46bisector will also pass through the
36:47center. This one will also pass through
36:49the center. So where will both lines
36:51cut each other? Both lines will cut
36:53each other at the center. As soon as
36:54the center comes, this gap is the
36:56radius. This gap is the radius. This
36:57gap is the radius. Once the center is
36:59found, the situation will become easy.
37:02Then everything will be solved easily.
37:04So, you will have to do two things
37:06first. Okay, son? You will have to do
37:08two things one by one. Now, let's begin
37:11. So first of all, one by one, I will
37:14find out. The equation of the
37:17perpendicular bisector of chord OP.
37:22Come on, here you go. You will have to
37:23do both things one by one. First, I
37:25will find this out. Let's name this
37:27equation number one. Come on, now where
37:30will it pass from? Where will the
37:31perpendicular bisector pass from?
37:33Through their midpoint. So what is the
37:35midpoint of these two? 0 + 3/2, here it
37:38is. And 0 + 9/2, here it is. So what is
37:42the answer to this? Okay? I added both
37:44of these and divided by 2. Here it is.
37:46So tell me, what will its midpoint be?
37:48Its midpoint will be 3/2, 9/2; here you
37:51go, this is the midpoint. The
37:53perpendicular bisector is a line itself
37:55, and for a line, you need a point and
37:57a gradient. Next step, if I ask you
38:00what the gradient of OP will be? Then
38:02you will say y2-y1 over x2-x1, so the
38:05gradient of OP came out to be three.
38:08The gradient of OP came out to be three
38:09. So if I ask you what the gradient of
38:11the perpendicular bisector will be? If
38:13I ask you what the gradient of the
38:15perpendicular bisector of OP will be?
38:17What will be the gradient of the
38:19perpendicular bisector of OP? You will
38:22say -1/3, the negative reciprocal, here
38:25you go. So this means that the gradient
38:28of this came out to be -1/3, here you
38:30go. Clear, son? This is the gradient.
38:33Now I need its equation. y-y 1 equals m
38:37times x-x 1, take this. y-y 1 equals m
38:43times x-x 1, take this; y-y 1 equals m
38:47times x-x 1. Now simplify it a bit more
38:51, simplifying gives -1/3x + 1/2 + 9/2 +
38:5510/2. That’s the equation of the
38:59perpendicular bisector. Clear, kid? So,
39:03half your work is done. One of your
39:05targets is achieved. One perpendicular
39:07bisector is found. Right? Now,
39:09obviously, a little bit of time will be
39:12wasted. We will have to work a bit
39:14slowly, obviously. Right? Okay, now you
39:17have another task. Now what do you need
39:19? Now, one more task. Now you need the
39:22equation of the perpendicular bisector
39:25of the one that remains. Of PQ, there
39:29you go. We need one more perpendicular
39:31bisector B. Meaning, the first one is
39:33done. Now we need the second. We need
39:35this now. Right? Now we have to find
39:37this out. Right? Right? Okay, let’s
39:40name this equation number two. Let’s
39:42name this equation number two. Now you
39:44have to find this second one out. Clear
39:46, kid? Now I need this. This is
39:47equation number one. So, you have it.
39:49Here you go. Now you will need equation
39:51number two. Here you go. Equation
39:53number one is here. Done. Now I need
39:55equation number two. Where will the
39:57perpendicular bisector of PQ pass
39:58through? The midpoint. So what is its
40:00midpoint? First, we have to find its
40:02midpoint. What is its midpoint? Its
40:05midpoint is 3 + 11/2, here it is. And 9
40:09+ 11/2, here it is. So now tell me,
40:13what is its midpoint? Its midpoint is
40:1614/2 which is 7, and 20/2 which is 10.
40:19So you get 7, 10. Here it is, 7, 10.
40:23Secondly, if you see, what is the
40:25gradient of PQ? Gradient of PQ is 11-9
40:29= 2/11-3 = 8, so its gradient is 1/4.
40:33Here it is, its gradient is found. So
40:35what will be the gradient of the
40:37perpendicular bisector? The gradient of
40:40the perpendicular bisector will be -4.
40:43The gradient of the perpendicular
40:45bisector will be -4. Here it is. Now we
40:48need its equation. We need its equation
40:51. So what will its equation be? y-y 1
40:54equals m times x-x 1, here it is. So
40:58what is its equation? -4 x + 28 + 10 =
41:01+38. This is equation number two as
41:05well. Now what do we have to do with
41:08these two equations? Tell me, we have
41:10to solve them simultaneously. Clear,
41:12kid? What do we have to do? Solve
41:14simultaneously. Now what will you have
41:16to do? Now you have to solve it.
41:18Simultaneously. Which ones? One and two
41:21. What do we need to do with one and
41:23two? We need to solve them
41:24simultaneously. Okay? Okay? It takes a
41:26bit of extra effort; it’s coordinate
41:28geometry but with circle properties.
41:30Now do one thing, take the y from here
41:32and put it in the equation above. Okay?
41:34Or let's just say, equate both the y
41:36values. Use whatever words you prefer.
41:39It's not a very complex logical point.
41:40Here it is. Now rearrange this. Here it
41:44is 4x-x/3, right here. What is 38-5? It
41:48is 33. 3*4 = 12, wait. One minute. Okay
42:08, let's see, 11x/3, when simplified
42:12further, it becomes 11/3x = 33. 11*3,
42:19so from this, we found that the answer
42:20for x is nine. There you go. When the
42:22answer for x is nine, just put this
42:25nine anywhere. 38-36 is 2, so the
42:27answer for y is two. So what did we do?
42:30Solved it simultaneously. When we
42:32solved it simultaneously, we got the
42:35values of x and y. So from this, I
42:37finally found out what the coordinates
42:39of the center are. 9, 2; this confirms
42:42that the center of my circle is 9, 2.
42:45There you go. This is the center of my
42:48circle, 9, 2, so my main target is
42:50achieved because once the center is
42:52found, the rest of the work will be
42:54easy. The rest of the work will now be
42:56easy. Okay, kid? Just as soon as the
42:57center is found, there is nothing
42:59extraordinary left for me to do. Here
43:01it is. So I have the coordinates of the
43:03center. Done, here it is. Okay? My
43:05center is found. What do I need?
43:07Equation of the circle. How will the
43:09radius come? OC is radius, PC is radius
43:12, CQ is radius. So if I ask you, how
43:15will the radius come? Radius? If I ask
43:17you, how will the radius come? The
43:19radius will come from the distance
43:21between O and C, that is also the
43:22radius. The distance between P and the
43:24center is also the radius. The distance
43:26between Q and the center is also the
43:27radius. So I checked, what is the
43:30answer for OC? 9 squared, 2 squared. So
43:3481 + 4, its answer comes to 85. So what
43:39is its radius? 85, there you go, the
43:42radius is found. Now I know what the
43:45radius of my circle is. The radius is
43:4885, and the center of the circle is
43:50also found. What is the radius? The
43:52radius is 85. There you go, the center
43:54is found and the radius is found too.
43:56There you go, the center is found and
43:57the radius is found too. Center plus
44:00radius. Both tasks are done. Clear, son
44:03? We have the center, and we have the
44:04radius. As soon as we have both the
44:06center and radius. Now the last step is
44:08the equation. So what will the equation
44:10be? We already know the equation. We
44:12have only studied one equation so far.
44:14(x-9) ², next (y-2) ², and what will
44:17be the square of √85? Its square will
44:21be 85. This is the derived final answer
44:25to your question. Alright, son? So,
44:28undoubtedly, this was a good question.
44:31The same type of question has appeared
44:32in past papers for five or six marks.
44:34In the same style, a question of the
44:36same style appeared. In past papers,
44:38for 5 to 6 marks. So, our target has
44:40been achieved. What was the target? The
44:42target was the equation of the circle.
44:43Done, son. Let's try more questions.
44:46Let's go. Finally, today we practiced
44:48more on the standard equation of the
44:49circle. Done, son? Now we have finally
44:52completed only one of the five points
44:55of this circle. Done, son? And what was
44:57the name of that one point? Standard
44:59equation of the circle. From the next
45:01task, God willing, we will start its
45:03second and third points. Okay? Allah
45:05Hafiz.

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