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AS Level Mathematics 9709 | Coordinate Geometry 12 | Relation between Line & Circle | Raja Shafqat — Transcript

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0:09Assalamu Alaikum, kids. Finally, today
0:12we will do two more points on circles
0:14so that circles are completed, and then
0:16in the next four or five classes, we
0:18will only practice circles from past
0:20papers. Today we will cover enough so
0:22that the basic concepts of the circle
0:24are finally complete, and after that,
0:26there will only be practice. Alright,
0:28the fourth point we are doing today is
0:30titled Relationship between Line and
0:32Circle. Actually, by the way, this
0:35point is a carry-forward. We studied
0:38another topic, another chapter. The
0:40name of that chapter was Quadratics.
0:43How is it related? Let's do one thing
0:46first: let's look at what we studied in
0:49Quadratics. Then I will show you how to
0:52connect it with today's topic. Actually
0:55, there are quite a few questions on
0:57this in past papers. When we studied a
0:58chapter, the name of the chapter was
1:00Quadratics. So, there was a point in
1:02Quadratics. The name of the point was
1:04Relationship between Line and Curve.
1:07The point we studied in Quadratics was
1:12named Relationship between Line and
1:16Curve. Relationship between Line and
1:20Curve. So, first, let me revise the
1:22basics a little bit on what concept we
1:24studied in Quadratics. Number one, in
1:27Quadratics, we learned that suppose we
1:30have a line. The equation of the line
1:34is mx + c, and suppose we have a curve.
1:37The curve is px² + qx + r, here it is.
1:41This is the same concept. Let me revise
1:43it a little for you first. We learned
1:45that if you have a line and a curve and
1:47you need to check the relationship to
1:49see what possible relationships can
1:50exist between the line and the curve.
1:53So, we studied four conditions in this.
1:56Four conditions. Condition number one
1:59we studied was if a line cuts a curve
2:01at two points. If the line intersects
2:04the curve at two different points. So,
2:07what condition did we study? That the
2:09discriminant is always positive. The
2:12discriminant is always positive. I am
2:14telling you this. The name of the
2:16chapter was Quadratics. So, we kept
2:18saying that whenever a line cuts a
2:20circle at two different points, what
2:22wording did we use? What term did we
2:25use? We kept saying b²-4ac, meaning
2:28the discriminant is always positive. So
2:30, what word or term did we use? We kept
2:33saying b²-4ac is > 0. Meaning we kept
2:37saying, what will happen in this
2:38situation? There are two distinct
2:41points of intersection. Meaning two
2:44different points of intersection are
2:46possible. Meaning if a line intersects
2:49a curve at two different points, the
2:51discriminant is always positive. This
2:53is what is being discussed. The name of
2:54the chapter was Quadratics. Then I will
2:56connect this to today's topic. Then we
2:58studied another condition. Then we
3:00studied another condition, that if the
3:03line is tangent to the curve. Like this
3:05situation here. Well, whether it is a
3:09happy face or a sad face. Right? Right
3:11now, our interest is not in the happy
3:13face or sad face. Our interest is what
3:15basic condition we studied. The first
3:17condition was that if the line cuts the
3:19curve at two points, then we used to
3:21say the discriminant is always positive
3:22. What was the second condition? If the
3:25line is tangent to the curve. Then we
3:27used to say that the discriminant is
3:30always zero. Meaning b²-4ac is always
3:33zero. b²-4ac will always be zero. So
3:37we used to say that the line is tangent
3:40to the curve. So what term or word did
3:43we use? We used to say the line is
3:46tangent to the curve. Meaning there is
3:50only one point of intersection. So, we
3:53used to read it like this. There is
3:55only one point of intersection. There
3:58is only one point of intersection. Then
4:00we used to study the third condition.
4:03The third condition was that if the
4:05line does not intersect the curve,
4:07meaning if the line does not intersect
4:09the curve, then we said the
4:10discriminant is less than zero. The
4:13discriminant is less than zero. When
4:15was this situation possible? When there
4:17was no relationship between the line
4:19and the curve. We used to say that in
4:21this situation, there is no point of
4:22intersection. We used to say no point
4:25of intersection. Meaning there is no
4:28relationship between the line and the
4:30curve. There is no relationship between
4:33, there is no relationship between the
4:36line and the curve. There is no
4:39relationship between the line and the
4:41curve. Clear? No point of intersection.
4:44Then we used to hear another word. That
4:46word was, sometimes it was written that
4:48the line meets the curve. Or sometimes
4:50the examiner said show that the line
4:52intersects a curve. So when this word
4:54is used, that the line meets the curve
4:56or the line intersects a curve. Then in
4:58this condition, we used to say that the
5:00discriminant can be equal to or greater
5:01than zero. We also used to say real
5:04roots. We also used to say real roots.
5:07Real roots mean a point of intersection
5:09is possible. What are real numbers? All
5:11numbers are real numbers. All numbers
5:13are real numbers. We have already
5:14studied all this previously. Meaning
5:16the point of intersection can be two or
5:19it can be one. Meaning if we combine
5:21both previous conditions, this
5:23condition will be derived. Look at this
5:25. If we combine the previous conditions
5:27, then this condition will be derived.
5:29Then we used to say that the line meets
5:31the curve at two different points or at
5:33one point. So, there used to be both
5:36conditions in this. Well, for those of
5:38you who remember this whole topic, we
5:40used to follow one method throughout
5:42this topic. Exactly the same method we
5:45will follow today. Exactly the same
5:47work we will do today. Okay? Which we
5:50used to do before this. We will follow
5:52the same method. What was the method?
5:55First, you look at the method. Whenever
5:57we solve a line and a curve, what do we
5:59find out? Point of intersection. So,
6:01what did we do throughout this topic?
6:04In this entire topic, we used to
6:06substitute y from the line into the
6:09curve. What did we use to do? We used
6:13to substitute y from the line into the
6:17curve. Into the curve. We have always
6:20studied this previous method that
6:22whenever this situation occurred, what
6:25did we do? We used to substitute y from
6:27the line into the curve, into the curve
6:29. Okay? First, let me revise this
6:31condition for you. Then obviously, I
6:33will show you how to prove it. Okay? We
6:36used to substitute y from the line into
6:39the curve. Why did we do that? Because
6:41sometimes the curve equation also has a
6:44y squared or an x squared term. Curve
6:47means the highest power is two. So, x
6:49could be to the power of two, and y
6:51could also be to the power of two. Or
6:53the curve could be in a form where you
6:55might have some trouble making a
6:57subject. So, the reliable option was
6:59that we keep substituting y from the
7:01line into the curve. So, as soon as we
7:03substituted y from the line into the
7:05curve, what was always derived? What
7:07was always found? A quadratic equation
7:10was always found. A quadratic equation
7:13was always found. Whenever that
7:16quadratic equation was found. Then what
7:19did we apply to that quadratic equation
7:21? We used to apply the discriminant.
7:24Then we used to apply the discriminant
7:27to it. Now, either a hint for the
7:29discriminant was given, or we had to
7:32prove something. Clear, dear? This was
7:34the method in the chapter we studied
7:36before, Quadratics, where the topic
7:38name was Relationship between Line and
7:40Curve. The topic today, now look at its
7:43name. Relationship between Line and
7:46Circle. Line and Circle. Okay? So now
7:49look at what the condition will be
7:50today. Today the entry will be such
7:53that you are given a line and you have
7:55to show that the line intersects the
7:57circle at two different points. So,
8:00math says if a line intersects a circle
8:02at two different points, the
8:04discriminant is always greater than
8:06zero. Meaning, what was it in that case
8:09? What was it there? There it was a
8:11curve. Today it is a circle. In those
8:13quadratics, we studied the curve. Today
8:15we are studying the circle. Otherwise,
8:17the method is 100%the same. Meaning, if
8:19a line intersects a circle at two
8:21different points. If a line intersects
8:23a circle at two different points. Then
8:25what do we say in the language of math?
8:27We say in the language of math that the
8:29discriminant is always positive. Then
8:31we studied one more thing. If, if math
8:34says if the line is tangent to the
8:36circle, meaning what was it here? Here
8:39it was two points of intersection. What
8:42was it here? Two distinct points of
8:44intersection. Is it clear, child? When
8:47the line cuts the curve at two
8:48different points, then two different
8:50points of intersection are possible.
8:52And what is the condition here? The
8:54condition here is that if the line is
8:56tangent to the circle, math says the
8:58discriminant is always zero. If the
9:02line is tangent to the circle, then the
9:04discriminant is always zero. Then what
9:07did we used to say? We used to say the
9:09discriminant is always zero. We have
9:11already studied this condition in
9:13quadratics. So, we kept saying that
9:15there is only one point of intersection
9:17. There is only, there is only one
9:19point of intersection. There is only
9:22one point of intersection. So, what
9:25word did we used to use? We kept saying
9:29that the line is tangent to the curve.
9:32Today we will say tangent to the circle
9:34. What was the third condition? If a
9:37line doesn't even touch a circle,
9:39meaning no relationship. Meaning, if
9:41the line does not intersect the curve,
9:43then what did we do at that time? If
9:45the line does not intersect the curve,
9:47then the discriminant is less than zero
9:48. So, here too the condition is the
9:50same, that if your line doesn't cut the
9:52circle at all. Meaning, if there is no
9:54point of intersection. If there is no
9:56point of intersection, then we kept
9:59saying the discriminant is always
10:01negative. The discriminant is always
10:03negative. And if the condition were
10:06just written like that, where the
10:07examiner says to show that the line
10:09meets the circle or the line intersects
10:11a circle. If the examiner says to show
10:14that the line meets a circle or the
10:16line intersects a circle. Then there
10:18can be two possibilities in this. Maybe
10:20the line cuts at two points or maybe
10:22the line cuts at one point. Meaning,
10:24both possibilities can happen. Just
10:26like we did in quadratics and the
10:28examiner said to show that the line
10:29meets a curve. So our working justified
10:32that "meets a curve" means two points
10:34of intersection or one point of
10:36intersection. So in this condition,
10:38what will we say? Just like quadratics.
10:40Like quadratics, we will say the
10:42discriminant is equal to or greater
10:44than zero. So when the statement "line
10:48meets the curve" is used, it means it
10:50will have both possibilities. It will
10:53have both possibilities. All right? If
10:55the line does not cut the curve at all,
10:57then both possibilities will exist. The
10:59method is also the same as the one we
11:01studied before in quadratics. All right
11:03? And I will show you how to apply it.
11:05No, it is exactly the same thing. The
11:07method that was in quadratics, use
11:09exactly the same method in this topic
11:11too. Meaning there is no difference in
11:13the method. In quadratics, we used to
11:15say what will we do? What did we do
11:17from the line? From the line, we used
11:20to substitute y into the curve. From
11:23the line, we used to substitute y into
11:26the curve; today we will say we
11:28substitute y into the circle. Here it
11:31was into the curve, here it is into the
11:33circle. Into the circle, so just like
11:36quadratics, what will be derived? Just
11:40like quadratics, you will derive a
11:42quadratic equation; you will derive a
11:44quadratic equation. Just like
11:47quadratics, a quadratic equation will
11:49be derived. Then what will we apply to
11:53that quadratic equation? We will apply
11:55the discriminant to that quadratic
11:57equation. Meaning the method is exactly
12:00the same. The method is exactly the
12:02same. So then what did we apply to it?
12:04Discriminant. Discriminant. Clear, son?
12:07And then there were the same two styles
12:08of questions. Either we had to show
12:11something, or we were given a hint and
12:14we had to find a constant. Clear, son?
12:17So this topic is almost the same. But
12:19what is the difference? Earlier it was
12:21a curve, today it is a circle. By the
12:23way, the same topic, same conditions,
12:26and same terms will be used in this.
12:28Clear, son? The paper pattern will be
12:31exactly like it was in quadratics. Now
12:33come, let me show you an example so
12:34that you remember. This is exactly the
12:36same thing. Clear? Come, let's start.
12:39Just look at the question. What
12:40question? Show that. The wording here
12:43is "show that." We had studied that
12:45exam-style questions are of two types.
12:48We studied before in quadratics as well
12:50, if we just explain what type of
12:52questions are in our syllabus. So
12:54exam-style questions were of two types.
12:56The first question we call questions
12:58with a given hint. When the examiner
13:00gave a hint, he would make you find
13:02some constant or another. It used to
13:04make us find the constant. Then the
13:06second style was that we had to show or
13:09prove something. Whatever the given
13:11statement was. The examiner would say
13:13show that the line intersects the curve
13:14at two different points. Show that the
13:16line is tangent to the circle, to the
13:19curve. Show that the line does not
13:21intersect the curve. So, we will do
13:23exactly the same thing here that we did
13:25in quadratics. Now, read the question.
13:28It is written here, "show that." So,
13:29let me explain it to you. Then we will
13:31practice more of this. I am from past
13:33papers. Look, if you check, what is the
13:35line equation given to you? This is
13:37your line. If you rearrange the
13:38equation of the line, your line is
13:40something like this. The line is 6-3 x;
13:42this is your line. This is your line,
13:45by the way. Right? It is clearly given
13:47in the question that this is the line,
13:49and the circle's equation is also
13:51clearly given. This is the circle's
13:52equation. x² + y² + 4x + 16y + 28 = 0
13:57, meaning the circle's equation is not
14:01given in standard form. It is given in
14:05expanded form. If anyone remembers, the
14:08standard form of a circle is what we
14:10used many times in this whole topic.
14:13The standard form was always this: it's
14:16always (x-a) ² and it's always (y-b)
14:19² = radius squared. By the way, we
14:22don't require this form. We are not
14:23interested in completing the square. We
14:25don't even need the center or the
14:27radius of the circle. We only need to
14:29show that the line is tangent to the
14:31circle. So, if I summarize, what is my
14:34target, by the way? My target is this.
14:37I have to prove that the given line is
14:40tangent to the circle. Meaning that
14:43line cuts the circle at one point. This
14:45is our target. We have to prove and
14:47show that, yes, the line and the curve
14:49are indeed cutting at one point. This
14:50is our target. Is it clear, kid? And
14:53our target is not even to find this
14:56point. To find the point, obviously, we
14:58would have to solve it simultaneously
15:00in full. Our target is just to show
15:02this. Our target is just to show that,
15:04yes, the line and the curve are cutting
15:06at one point. Meaning, we will take
15:09help from the discriminant. Now, see
15:11what we will do. From the line, plug y
15:13into the curve, into the circle. I know
15:16this is our circle. x² + y² + 4x +
15:2016y + 28 = 0. This is the same style,
15:24the same method we did before in
15:28quadratics. Alright, now what will we
15:31do? We will replace y with y. What is y
15:34? ys x-cx, replaced y with y. Is that
15:38okay, son? Wherever y is involved, I
15:40replaced y with y. Look, I have
15:42replaced y with y. Clear, son? Now,
15:45what kind of equation should be derived
15:47? Quadratic. Now, simplify this
15:49equation a bit. Expand it. This will
15:52become a²-2*a*b and + b², there, I
15:56expanded it. Clear, son? Now 16*6 = 96,
16:00I expanded this too. 16*3 = 48, here it
16:03is, and + 28 = 0, we simplified it
16:06further. Okay? Now, what do 9 and 1
16:08become? Simplified it further. This
16:10became 10 x², and finally, what do you
16:12have? Finally, you have -36 and -48,
16:17that is -84, there, -84 x is here, and
16:22plus 36, 96, and 28, I added these
16:26three, it came to plus 160. There, it
16:30became 160. As always, I did the same
16:33work as we keep doing in quadratics. We
16:36know if I take something common from
16:38this, wait, hold on a minute, son. Yes,
16:42by mistake, I missed 4x here. Exactly,
16:47I missed the 4x. So, I will add 4x.
16:50Here it is, look, I had missed 4x. So,
16:54when I added 4, what is the final
16:56answer? The final answer is -80. There.
16:59Okay? Now the correction is done. Now
17:01rearrange it. Take 10 as common. This
17:04became 10 (x²-8x + 16) = 0, the
17:06equation is derived in this form. What
17:08kind of equation was derived? A
17:10quadratic equation was derived. So, we
17:12know this is 100%method-based work. We
17:14put y from the line into the curve.
17:16Today we will say into the circle. A
17:18quadratic equation will always be
17:20derived. Now, what is your target? To
17:23show that the line is tangent to the
17:25circle. Then what will we apply? Now we
17:28will apply the discriminant. What is
17:30the formula for the discriminant? b²-
17:324ac, here it is. So, the discriminant
17:36formula is b²-4*a*c, so when I
17:39simplify it, it becomes 64-64, so it is
17:43confirmed that the discriminant is zero
17:47. This was your final target. This is
17:50what you had to show. Clear, son? This
17:52is what you had to show, that the line
17:54is tangent to the circle. Now I will
17:56show you an extra task. No doubt, it is
17:58not required. By the way. I am showing
18:00you extra work. It is not required.
18:02That's it, what needed to be proved is
18:03done. Now, if the examiner asks me to
18:06find the point of intersection or the
18:09point of tangency, meaning the point
18:12where the line cuts the circle, it's
18:15not an issue. We know that when we
18:18solved them simultaneously, this
18:20equation was derived. Obviously,
18:23solving simultaneously led to the
18:24equation. What was the equation you got
18:26? Just look, you solved it
18:28simultaneously. So, this is the
18:30equation you got. Right? Okay? So you
18:32proved it with the help of the
18:33discriminant. That's it. It is proven.
18:35Now, what should you do with this?
18:36Solve it further. We know what solving
18:38simultaneously helps us find, right?
18:40The point of intersection. So, from
18:42here we found that both values of x are
18:44what? The same. In quadratics, we also
18:46used to call this a repeated root. Okay
18:49? One point of intersection. So, only
18:51one value of x came from here. What did
18:53that x value become? It became four.
18:54What did that value of x become? The
18:56value of x was found to be four. Now,
18:59if you put this four into y, what is
19:02your y? Your y was 6-3 x, that was your
19:05y. Now put four into this. So 6-3 times
19:094, so what is the answer for y here?
19:12The answer for y is -6. So from here I
19:15found that my point of intersection, or
19:17the point of tangency, what were the
19:19coordinates of that point? The
19:22coordinates came out to be 4 -6. I know
19:26very well that this is not even
19:28required in the exam. It is not written
19:30anywhere in the question that you have
19:31to do this or find this out. I just did
19:34some extra work so that if the examiner
19:37asked to find the point of tangency, we
19:40could have done one more extra step and
19:43found the point of tangency. So that
19:45means this point of tangency, what
19:47would this point of tangency be? This
19:51point of tangency would be 4 -6, that
19:54is, this is your line. Your line says
19:586-3 x, right? And this is your curve.
20:00By the way, this is your curve, meaning
20:02this is your circle. Is this clear,
20:04dear? So it is not required by him, the
20:06examiner didn't ask us to find the
20:08point of tangency, but there are two or
20:10three questions in past papers where
20:12the point of tangency came up, and the
20:14examiner asked to find the point of
20:16tangency. It has appeared in the past
20:18papers, is this clear, son? That is why
20:19I got it completed. So when I provide
20:21more practice from past papers, this
20:23method-based topic will frequently keep
20:25repeating. Clear, son? Finally, the
20:27last point of circles, so that circles
20:30are completed first. Alright, son? Come
20:32on, now look, what is this last point
20:34called? Finally, this is the fifth
20:35point. The name of this fifth point is
20:37the General Equation of a Circle. Now I
20:39am telling you something new. Now
20:41please focus and watch carefully. Look,
20:44what did you study previously? I will
20:46explain this in detail again. The
20:48equation of the circle that you and I
20:50have studied so far, what was it?
20:52Previously, we studied this equation.
20:54What did we call this equation in the
20:56language of math? We used to call it
20:59the Standard Equation of a Circle.
21:01Previously, we used this equation many
21:03times. The name of this equation was
21:06the Standard Equation of the Circle.
21:09First, let's look at what we studied
21:10previously. Then we see what the new
21:12thing is. We have studied many times
21:14that the standard equation of a circle
21:16is this. We studied that this is called
21:20the standard equation of a circle: (x-a
21:22) ² Plus, then we studied (y-b) ²
21:27here, and what is it equal to? r²,
21:30this is what we studied previously. We
21:33know that in this situation, we kept
21:34saying what the center of this circle
21:36would be. a, b and then we kept saying
21:38what the radius of this circle would be
21:40. The radius of this circle would be r.
21:43Then we kept saying, what are x and y
21:45in this? We used to say that x and y
21:48are...Any point on the circle, every
21:50point on the circle, or you could say
21:52any point on the circle. Any point on
21:56the circle. This is what we studied
21:59previously before this. Done, son.
22:02Previously, we studied this. If I show
22:04you its graph, technically we used to
22:06say this means this is the center of
22:08the circle. This is the center of the
22:11circle. Alright, son? We used to use it
22:13like this. We used to say this is the
22:15circle's center and this is its radius.
22:16I am repeating the old work on purpose.
22:19Then I will show you the new work. This
22:20is its radius. We used to call the
22:23coordinates of the center a, b, and
22:25whatever point is on the circumference,
22:28we call it x and y. In this way, we had
22:30seen the equation derived. This is the
22:32old work. Today we are doing a new task
22:34. Alright, son? What is the new task?
22:36First, listen. Then, if you are
22:37interested in the logic, I will explain
22:39its entire logic. Math says, math says
22:41today we are studying something new.
22:43What is the name of this new thing? The
22:44name of this new thing is the General
22:45Equation of a Circle. This is the new
22:47thing. The name of this new thing is
22:49the General Equation of a Circle. Now
22:51what is this? Where was it derived from
22:53? I will guide you through it. General
22:55Equation of a Circle. Then we also see
22:57why we are studying this. Why are we
23:00studying this? The General Equation of
23:01a Circle, what is this thing? Math says
23:03that if you are given any equation in
23:07this form: x² + y² + 2gx + 2fy + c =
23:110, then this equation is read as the
23:15General Equation of a Circle. I will
23:18guide you. I will explain the whole
23:20logic. General Equation of a Circle.
23:22Math says if you have a question in
23:23this form, in the form of the General
23:25Equation of a Circle, then what will be
23:27the center of this circle?-g, -f; then
23:30we see why we are studying this. -g, -f
23:33will be the center of this circle. And
23:37if you are interested in the radius,
23:38then what will be the radius of this
23:40circle? The radius of this circle will
23:42be this. The radius will be √g² +
23:45f²-c; in the language of math, this
23:48base concept is called the General
23:50Equation of a Circle. Now what is this?
23:54If any student is interested in the
23:55logic, I will guide you in detail. Okay
23:57, what is the plus point of this method
23:59? Why do we use the general equation?
24:02What do we do in the general equation?
24:04What do we do in the general equation?
24:07In the general equation, without
24:10completing the square, without
24:13completing the square, without
24:16completing the square, what do we do?
24:21We find out the coordinates of the
24:23center. Without, without completing the
24:26square, we find out the center. Without
24:28completing the square, we find out the
24:31radius. Without completing the square.
24:33This is its biggest plus point. This is
24:35an alternate method. Basically, what is
24:38this? This is an alternate method. An
24:40alternate method for what? This is an
24:42alternate method to completing the
24:43square. But the most interesting thing
24:45is that it is an alternate method to
24:47completing the square. Where was it
24:49itself derived from? It was itself
24:50derived from completing the square. Now
24:53, I will show you its logic in detail.
24:55Then I will have you practice it. Now,
24:57suppose any student has a doubt about
24:59my work. Come, let's do one thing. I
25:01will show you the proof of this. Okay?
25:05If any student has a doubt, then come,
25:06let me show you the proof of this.
25:08Watch carefully, son. I told you today
25:11that this equation, x² + y² + 2gx +
25:162fy + c = 0, is called the general
25:21equation of a circle. Okay. Now,
25:25suppose I have a doubt about its center
25:28and radius. Suppose I have a doubt. So,
25:31I will show you its proof. To see the
25:33proof, we will need help from
25:35completing the square. The interesting
25:38thing is x² + 2gx, what did I do? I
25:42copied the x terms. Is that okay, kid?
25:45Just like we used to do in completing
25:46the square, we used to combine them.
25:48Similarly, I will combine the y terms
25:51as well. Here it is, y² + 2fy, I have
25:54combined the y terms too. Okay? And + c
25:57= what did it become? Zero. Here you go
25:59. We rearranged the x terms first
26:01because we know I am converting this. I
26:04will show you by converting this into
26:06standard form so you get an idea and
26:08realize that the center and radius are
26:10truly the same. This is it. Now watch
26:13what I am doing. Now I am applying
26:14completing the square. Watch my working
26:17carefully. Now I am applying the
26:19completing the square method. Here you
26:21go. You don't have to do this method in
26:24the exam. But I am telling you this
26:25first to explain the logic. I am
26:27applying the completing the square
26:29method. Is that okay, kid? Completing
26:33the square method. Let's check. Now
26:35apply completing the square to this
26:36first. This became a squared. In the
26:39center is 2*a*b. So obviously we know
26:42how b used to be found. b was the
26:44coefficient divided by 2, this is your
26:47b. So if I know completing the square,
26:50then I know that its b is g. By the way
26:52. So that means I will add g squared
26:55and also subtract g squared. So I have
26:57applied completing the square to the x
27:00terms. Similarly, now apply completing
27:03the square to the y term as well. How
27:06will I apply it? I said y squared, the
27:09center is 2*a*b. So how does b come?
27:12Coefficient divided by 2, so the
27:13coefficient is 2f. Divide it by 2. What
27:16was the coefficient with x here? So do
27:192g/2. So the b that you got here is
27:22basically what? It came out to be f. So
27:24then we will add f squared and subtract
27:27it as well. Done, kid. We have added f
27:30squared and subtracted it as well.
27:32Clear? Added f squared as well. It is
27:35exactly the same task we used to see in
27:37completing the square before. Okay, son
27:40? I am doing the same thing. So, I
27:42haven't done anything new. Okay? Look,
27:44I applied completing the square on this
27:46too. And copy C as it is. There you go.
27:48Copy C as it is. I am only showing this
27:51to explain the logic to you. Look here.
27:53Now, if you know completing the square,
27:55then this whole thing is complete. If
27:57you know completing the square, then
27:59this whole thing is complete. What will
28:01you do now? Now convert this into the
28:03perfect square form. So, what did this
28:05become after converting to the perfect
28:07square form? x + g squared, and copy-g
28:10squared as is. Similarly, what did this
28:13become? This is a squared + 2ab + b
28:16squared, so what did this whole thing
28:18become? y + f squared, copy-f squared
28:21as is, and copy c as is. Here is the
28:25last and final step. Now arrange this a
28:28little more. This became x + g squared.
28:31This became y + f squared, and shift
28:33the remaining terms to the right side.
28:36g squared + f squared-c, here it is.
28:39Now what did you do? You applied the
28:41completing the square method to the
28:44general form. Clear, son? Now compare
28:47this. Now compare this with the
28:50standard form. Compare this with the
28:54standard form, with the standard form.
28:57So when you compare it with the
28:59standard form, you'll say, this was the
29:02standard form: (x-a) squared + (y-b)
29:05squared = r squared. So when you match
29:09it with this, g becomes-a, so a becomes
29:12-g. From here, after comparing, I found
29:15out what a is. a becomes-g, here a is
29:18compared. Next, match y, here is y.
29:22When I compare from here, what is f? b.
29:25So I can say that b became what? b
29:27became-f. From here I found out that my
29:31b became what? b became-f. There you go
29:34, b is also here, b became-f. Your b is
29:36here, see, b is here too. Now compare
29:39it, so say what r squared became? r
29:41squared is equal to the radius squared,
29:44what is it equal to? Radius squared is
29:47g squared + f squared-c. Radius is
29:50always length, it is positive. So from
29:53here, I found out what the answer for
29:55radius r is? The answer for the radius
29:58is g² + f²-c, so from here I found
30:00out that this is the radius. This is
30:03the radius. Now you know what the
30:05center is? The center is always a, b.
30:09So from here, I found out what the
30:11center is? The center is-g and-f, here
30:13it is. I proved to you that it really
30:16is like this. The interesting thing is
30:19that this proof is not even required in
30:21the exam. So what new thing did we
30:23learn today, that if you have the
30:26circle's form in expanded form, meaning
30:29data isn't given in the whole square
30:31form? Data is not given in the x-a²
30:34y-b whole square form. Data is not
30:37given in this form. So what will you do
30:39? You will see that you will compare it
30:42with this equation. So today we learned
30:44a new equation. Its name is the General
30:46Equation of a Circle. General Equation
30:48of a Circle. This is a new thing we
30:49learned today. General Equation of a
30:51Circle. So from here, you found out
30:54what your c is? What is the center?-g,
30:56-f, what is the radius? g² + f²-c. So
31:01this is basically an alternate method.
31:03The interesting thing is that this is
31:04an alternate method. An alternate to
31:06what? The method of completing the
31:08square. This is an alternate to
31:09completing the square. So, what is this
31:12basically? You will compare the
31:14equation with this. So, this is
31:16basically an alternate method. I can
31:19say that the name of this method is the
31:22Comparing Coefficient Method. What
31:25should the name of this comparison
31:27method be? Comparing Coefficient Method
31:30. So, I told you the whole story of how
31:33this was derived. It's not required in
31:35the exam at all. By the way. Clear? For
31:38a student who doesn't know the
31:39completing the square method, this is
31:41the best option for them. Look how,
31:42child, I will show you in detail. Look
31:44here, I copied this summary here. Here
31:47you go, the same summary that I
31:49explained to you today, I copied here.
31:52Clear, child, today I will tell you
31:54what the plus point of this method is.
31:56Why did we study this method? Look, I
31:58will show you a small example so that
32:00you remember. Look here, if a question
32:02comes in the exam like this. Here you
32:03go, if a question like this comes in
32:05the exam, x², we have done this before
32:07, by the way. Right? We studied a topic
32:10, uh, the first day, Standard Equation
32:12of a Circle. We studied this thing at
32:14that time too. Look at this: x² + y²-
32:1724x-18y + 125 = 0. Suppose this
32:21question appears in an exam. This
32:25question appears in an exam. And the
32:27examiner is asking to find the center
32:31of the circle and find the radius of
32:34the circle. The examiner is asking to
32:37find the coordinates of the center and
32:39find the radius. Okay? This is the
32:41condition in the exam. Now I will
32:42revise the old topic for you and also
32:44show you new work. Option number one.
32:47Option number one. Look, a student who
32:51knows the standard form. What will a
32:55student who knows the standard form
32:57have to do? They will have to apply
32:59completing the square. Okay? Look, the
33:01student who needs to convert it into
33:04this particular form. So what will that
33:06student do? First, they will collect
33:08the x terms. We have already studied
33:09this. The name of this topic was
33:11standard form. Then what will the
33:13student do? They will collect the y
33:14terms. Here it is, and + 125 = 0 is
33:18here. Done. Then the student will apply
33:22completing the square. Then after 2
33:24minutes, I will tell you an alternate
33:26method for completing the square. Okay?
33:29For completing the square, meaning what
33:31will a student do who doesn't know
33:32completing the square? Now you complete
33:35this first. a squared, there is a minus
33:38sign in the center. So there will be a
33:39minus sign in the formula too.
33:40Obviously. So what was the formula? a²
33:43-2*a*b, we know how b was derived.
33:48Coefficient of x/2, so its b is 12.
33:50There you go, b is 12, so what will we
33:52do? We will add and subtract 12 squared
33:55. This method was called completing the
33:58square. Then we will apply the same
34:00work on y. On y, look here: y, 18/2 is
34:059. So what will be in the center? 2*a*b
34:07, here b is 9. So we will add and
34:10subtract 9 squared. Applied it, and
34:13copy + 125 as it is. Here you go, 125
34:16is copied. Then what did we learn in
34:19completing the square, that this whole
34:20thing is now complete. Similarly, this
34:23whole thing is also complete. This was
34:25the typical method, the method based on
34:27what was called completing the square.
34:29Completing the square. So today we
34:31learned a new method, an alternate
34:33method. I will tell you the details of
34:34that now. Now tell me, what did this
34:36whole thing become after completing?
34:38This whole thing became (x-12) ² and
34:41what is this? This is -144 copied as it
34:44is, right here. Plus, what is this
34:46whole thing? This whole thing becomes y
34:50-9² and -81 copied as is, and 125
34:54copied as is. Here you go, kid. This is
34:58the same old work. Now, if I make it
35:00easier, it becomes x-12². y-9² Right?
35:07And when we simplify this, it becomes
35:09144. 144 + 81-125, this is what we get.
35:16One minute, wait, kid. 144 + 81-125,
35:18simplifying this gives the answer 100,
35:22here it is, the answer is 100, clear,
35:25kid? Now, the student who knows the
35:28standard form will compare it with the
35:30standard form, so what will they say?
35:33They will say, what is 'a' for this
35:34question? 12, what is 'b'? Nine. What
35:37is the radius squared? 100, so from
35:40here we found that our 'a'—from here
35:43we found that our 'a' is 12 and what is
35:46our 'b'? 9 and our center is 12, 9 and
35:51the radius squared is 100. So, from
35:54here we found out what the answer for
35:55the radius is. The radius answer is 10.
35:57There you go. From here we found out
35:59what the answer for the radius is. It
36:01is 10. That's it. This was the old way.
36:03Clear, kid? We have already studied
36:05this. The topic name was Standard Form.
36:07Today we are doing new work. Now I will
36:09show you how to solve this question
36:11again. Today we studied a new method.
36:14Now we will use option number two.
36:17Option number two. We will use a new
36:19method. What is the name of this option
36:21number two? The name of this option
36:23number two is General Equation of a
36:25Circle. The name of this option number
36:28two is General Equation of a Circle.
36:30Done. This is quick. This is an
36:32alternate method of comparing
36:34coefficients. What method do we use in
36:37the general equation of a circle? What
36:40are we actually doing? In this, we are
36:43doing comparing coefficients. What do
36:46we actually do in this method? What do
36:48we do in this method? We compare
36:50coefficients. Meaning, this is an
36:52alternate method to completing the
36:53square. But what is the interesting
36:55thing? Where was it derived from? It
36:56was derived from completing the square.
36:58Look kid, now this is your question in
37:03the exam: x² + y²-24x-18y + 125 = 0,
37:09here it is. I need the center of this
37:12circle. So, what will I do? I will
37:14compare it with the general equation.
37:17x² + y² + 2gx + 2fy + c = 0, now when
37:23you match this, x² and y² cancel out.
37:29So, after comparing, you found that 2g
37:32is equal to -24. From this, you found
37:35out what g is? Your g is -12. Here, g
37:39is -12. Similarly, compare what your 2f
37:44is. Your 2f is -18, here it is. So,
37:46what is the answer for f from this? The
37:49answer for f is -9, here it is. And in
37:52the same way, if you compare, what is
37:55your c? Compare it, your c is 125, here
37:57it is. I did not apply any 'completing
37:59the square' method here. Clear, kid?
38:02That is why this method is called
38:03comparing coefficients. In mathematics,
38:06this method is called comparing
38:07coefficients. Comparing coefficients.
38:10Is that okay, kid? Comparing
38:11coefficients. Now look, I told you
38:13today that if a student uses this
38:15method, what will the center be?-g, -f,
38:17meaning change the sign of g, and-f,
38:20meaning change the sign of f as well.
38:24So from this, you found out what your
38:26center is? 12, 9, what did your center
38:28come out to be? 12, 9, and just one
38:30last thing, I told you what the formula
38:33to find the radius is. What is the
38:35radius formula? The radius formula is
38:39√g² + f²-c. Now, put the values in
38:43it. g² + f²-c, what is g²? g², and
38:47g is -12, here it is. g² + f²-c, here
38:54it is. You have put in c. What was your
38:57c? Your c was 125, here you go. Okay?
39:00By the way, all of this is inside the
39:01root. Here it is. Now when you simplify
39:04it, it becomes 100, and the root of 100
39:07is 10. The same radius was found, and
39:09the same center was found. Clear? This
39:13is the method. So match it, look, what
39:15is the radius? Same. Here is the center
39:18, what is it? Same. Here it is. This
39:20method is what we previously studied in
39:22quadratics, the 'completing the square
39:24method', and we also studied it in the
39:26first point of this chapter, which was
39:28named 'Standard Equation of a Circle'.
39:30So if someone asks us what the general
39:32equation of a circle is? What is this
39:35thing? This is the comparing
39:37coefficients method. This is an
39:39alternate method for those students, an
39:43alternate method for those students who
39:46do not know completing the square.
39:50Clear, kid? I will copy it here, what
39:52exactly is this thing? This method is
39:55an alternate method. Actually, this is
40:00an alternate method of completing the
40:02square. Well, I have written it here
40:05and even showed you a question to prove
40:07that it really is the case. I even
40:09proved the question for you to show
40:11that it really is the case. Is that
40:12clear, son? The interesting thing about
40:14the exam is this. There are no
40:16restrictions in the exam. There are no
40:18restrictions in the exam. Alright? So,
40:21I intentionally repeated an old method
40:23and also taught you something new. Is
40:25that clear, son? There is never any
40:26restriction in the exam. If a student
40:28says that they don't want to use the
40:30general equation form at all. Alright?
40:32Then don't use it. You can solve the
40:34entire chapter even without it. Without
40:36the general equation of a circle, your
40:38entire topic will be solved. But often,
40:40when you read the marking scheme,
40:42sometimes the examiner uses this method
40:44too. So it is best that you know every
40:47point of the syllabus. Alright? Then
40:49you can apply it according to the
40:51situation. Is that clear, son? Let me
40:53show you a couple more examples. So
40:55finally, I showed you this new method
40:57today. I told you something new today.
41:00Then I explained what this is. This is
41:02an alternate method. An alternate
41:04method to what? This is an alternate
41:06method to completing the square. And
41:07then I proved it for you. The
41:09interesting thing is that while it is
41:11an alternate method. But where was it
41:12derived from itself? It was derived
41:14from completing the square itself. How
41:16interesting is that. It is an alternate
41:19method of completing the square. But
41:20logically, where did it come from? It
41:22was derived itself. It was derived
41:24itself from completing the square.
41:27Alright, son? Let me show you one or
41:28two more examples of this. This was
41:29finally your last point. Is that clear,
41:31son? Let me show you one or two more
41:32examples. So there is never any
41:33restriction in the exam. Alright, son?
41:35Apply whichever one you feel like. Just
41:37like I showed you by solving a question
41:39using both methods. Now look, let me
41:40show you two or three more examples.
41:42Suppose this question comes in the exam
41:44. x² + y²-x-10y + 5 = 0, there it is.
41:51This question came up. Clear? And the
41:54examiner is asking me to find the
41:55coordinates of the center. Find the
41:58coordinates of the center and find the
42:01radius of the circle. Alright? Now I am
42:04solving this question without
42:05completing the square. Alright? Suppose
42:08I don't know how to complete the square
42:10. We have used completing the square
42:12many times. So today, I am solving it
42:14for you with a new method, without
42:16completing the square. So, the method I
42:18am using today, the name of that method
42:20is the General Equation of a Circle.
42:22What is the name of that new method?
42:24The name of this new method is the
42:26General Equation of a Circle. General
42:28Equation of a Circle. There you go, in
42:30the General Equation of a Circle, what
42:31do you just have to do? You just have
42:34to compare any given question with the
42:36general equation. Here, this is the
42:39general equation. You just have to
42:41compare it. Come, let's compare. x²,
42:43x², done. Okay? y², y², done. Now
42:46tell me, what is with x? 2g, and what
42:48is here? -1, so you have compared it.
42:51What did you compare? The coefficient
42:54of x. Okay? So you first compared x
42:56with x. There you go. I will mark here
42:59that you compared x with x. What is
43:02with x? 2g, and what is with x in your
43:05question? -1, here it is. So from here,
43:07we found out what the answer for g is.
43:09It is -1/2. There you go. From here,
43:12your g is found. There you go. From
43:14here, your g is found. That's it. Next,
43:17g is found from here. Just finished.
43:19It’s a matter of seconds. I didn't
43:21apply any completing the square method
43:23here. Next, now tell me, what is with y
43:25? -10, what is with y? 2f, so now in
43:27the same way, what will you do? Now you
43:30will compare the coefficient of y. Okay
43:33, kid? Now you will compare the
43:35coefficient of y. Compare the
43:38coefficient of y. There you go. Tell me
43:41, what is with y? In the question, it's
43:44-10 with y. In the formula, it's 2f
43:46with y. So, what is the answer for f
43:48from here? The answer for f is -5.
43:50There you go. f is found from here.
43:51Clear, kid? There you go. So, I am
43:54getting you to practice a little more.
43:56This is very method-based work. There
43:57you go. This is also done. What is the
44:00formula to find the radius? I mentioned
44:02the radius formula above. That is why I
44:03have marked it too. g² + f²-c, here
44:07it is. This is the formula to find the
44:09radius. Come on, now let's see what we
44:11will do. g squared is -1/2, squared
44:14becomes 1/4. f squared, if you square f
44:18, what will it be? -5 squared is 25,
44:21next is the last step. You look at what
44:23your c is. Your c is 5; when you
44:26compare, what is your c? It is 5. So
44:29what is c? Five here. Put five, clear?
44:32So what does this become, 20.25? So we
44:35checked the root of 20.25, okay. We
44:38checked the root of 20.25 and the
44:40answer is 9/2. There you go, your
44:43radius is 9/2. That’s it, the radius
44:45is found. Now tell me, what did we
44:47study today? What is the center? The
44:49center is-g, -f. There it is, so what
44:52is the center? The center is 1/2, -5;
44:55center is 1/2, -5. 1/2, -5, this is the
45:01center. What is the radius? The radius
45:04is 9/2; your target is achieved. Clear,
45:07kid? Your target is achieved. What was
45:09the target? Coordinates of the center.
45:11So let me summarize again. The general
45:14equation method is called comparing
45:15coefficients. What is the general
45:17equation method basically? The general
45:20equation method is an alternate method
45:22to solve questions without completing
45:24the square. So what is the general
45:25equation method? I will explain it
45:27again. What method do we use in the
45:29general equation method? We use
45:31comparing coefficients. We use
45:34comparing coefficients. This is
45:37actually the logic behind it. So we are
45:39always using comparing coefficients.
45:41Clear, kid? I will show you one or two
45:43more examples of this type. So what did
45:45I actually do in this? Just, just, just
45:48compared. Okay, kid? Today we learned
45:51an alternate method. Alternate. Whose
45:53alternate method is this? This is an
45:55alternate method for you. This is an
45:57alternate method for completing the
45:59square. Okay? I will show you one or
46:01two more questions. Previously we did a
46:03topic. In that, we learned the standard
46:05equation of a circle. I will pick one
46:06or two questions from that to show you.
46:08Let's try again. Look at this. Suppose
46:10you get a question in the exam. The
46:12question is 2 x², I am taking a
46:14question from the practice book. 2 x²
46:17+ 2 y² -4 x-10y-163 = 0, here it is.
46:25Come on, suppose this is a question.
46:27Okay? This is the question. I don't
46:29know, I don't know completing the
46:30square. Clear, kid? I don't know
46:32completing the square. Now I am telling
46:35you that what do I need? Without using
46:38completing the square. Without using
46:40completing the square. I need the
46:43center of the circle. Okay? Without
46:45using completing the square. What do I
46:47need to find out? I need to find out
46:49two things. Number one, I need to find
46:51the center of the circle. What is the
46:54center of the circle? Okay, I need to
46:56find out the center of the circle. The
46:59center of the circle, and I need to
47:01find the radius of the circle. Both the
47:04center and the radius. All right? So, I
47:06am using a new method today. The new
47:09method I mentioned today is called the
47:11General Equation of a Circle. So, I
47:14will use that method today. Because we
47:16have used completing the square many
47:18times. General Equation of a Circle.
47:19What is this method basically? It is
47:22comparing coefficients. What is this
47:24method basically? This method is
47:26basically comparing coefficients. This
47:29method is basically an alternative to
47:31completing the square. Come on, let's
47:33start now. Now look, what is your
47:36question first of all? 2 x² 2 y² -14
47:42x-10y-163 = 0, this is your question.
47:47Clear, kid? The issue is that in your
47:50form, there is always a one with x².
47:53So you will have to take two as a
47:54common factor. There you go, with a
47:56little extra effort you took two common
47:58. This became 2*7. This became 2*5 and
48:01this became 2*81.5. Now your equation
48:03has converted into a proper shape.
48:05Converting to a proper shape means that
48:08in the standard general form, x² and
48:11y² must always have a one. Now let's
48:13begin our question. Now we will compare
48:16it. There you go. Now what is your job?
48:18Your job is to compare it with the
48:20general equation. x² + y² + 2gx + 2fy
48:28+ c = 0, there you go. Now our job is
48:30to match it with this. Now look, match
48:33it. This has cancelled out. There is -7
48:35with x. Here there is 2g. Okay, kid? So
48:38now, first match x. Compare it. Now
48:41tell me, what is with x? With x, there
48:45is -7 and here there is 2g, so what is
48:47the answer for g from here? The answer
48:51for g from here is -3.5, this is your g
48:53. That's it. This is your g. Similarly,
48:57now you compare y. Coefficient of y,
49:00meaning compare y. Coefficient of y,
49:02compare y. Okay, kid? Now tell me what
49:05is with y? -10, what is here? 2f, 2f,
49:08so you compared it. Sorry, this was -5,
49:12with y is -5, there you go! With y is
49:14-5, here 2f. So what did the answer for
49:17f become? The answer for f became -5/2.
49:20You have found f as well, there you go!
49:23f has also been found. So, you already
49:25know what the remaining value is, so
49:26what is your c? Match it and tell me,
49:28what is c? c is your...c is your -81.5,
49:31that is your c. Now tell me, what is
49:35the formula for the radius? We studied
49:37the radius formula today. The radius
49:39formula is √g² + f²-c, here it is.
49:42Now, put the values into this. Okay,
49:45child? Now, put the values into this.
49:47g², right? Now, what will you do here?
49:50You just have to put the radius values
49:52in here. So now, what will you do? I
49:53will show you how to put all the values
49:55in for g. Here you go. I will show you
49:57by putting in all the values. So, how
49:59will the radius be found? You have to
50:01put the values into the formula. Here
50:04you go. Come on, put the values into
50:05the formula. What is your formula? Your
50:08formula is g² + f²-c, so what was c?
50:16c was -81, so minus-minus becomes what?
50:19+81.5, clear, child? Now, you just have
50:22to simplify this. 3.5² + 2.5² + 81.5,
50:27the square root of the answer is 100.
50:32So, tell me, what is the root of 100?
50:35The root of 100 is 10. Here you go, the
50:37root of 100 is 10. So, the radius came
50:39out to be 10 from here. So, I found out
50:41from here that the radius is 10.
50:42Alright, the radius has come. The
50:45radius has come. Today, we studied what
50:47the coordinates of the center are. The
50:49coordinates of the center are-g, -f;
50:51this is your center. -g, -f is your
50:55center. So now tell me, the center-g,
50:58that is, the opposite sign of g, so
51:00this becomes +3.5, and-f, that is, the
51:03opposite sign of f, this becomes 2.5,
51:05so your whole question is now solved.
51:09So now, if anyone asks us, what is this
51:11general equation of a circle? Then we
51:14will tell them it is an alternate
51:16method to completing the square. It is
51:18an alternate method to completing the
51:20square, and what is found with its help
51:21? With its help, we find the center of
51:23the circle and the radius of the circle
51:25. Done, child? Today, finally, all the
51:28concepts of the circle have been
51:29covered. Now, we will only and only
51:31practice it overall. Insha'Allah, we
51:34will practice this continuously for the
51:36next three or four classes. Okay, child
51:38? Allah Hafiz.

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