MATHS GRADE12 ASSIGNMENT 2026

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[00:00:02]Hello grade 12s. Welcome to academic world. Let's continue with your assignment. We are looking now at question 3. 3.3.1.

[00:00:15]Let's see what they want on this question of 3.3.1.

[00:00:20]I want to select a pen. Then we are ready to go.

[00:00:25]Let me select a pen. Yep.

[00:00:31]Okay. The pen it's selected.

[00:00:35]Okay. Yes. So they are saying prove that angle t is equals to 180 - 2x. This is the first question. So guys in this case O is the center of the circle.

[00:00:55]O is the center of this uh smaller circle.

[00:01:01]Okay. So I know that the angle at the center in this case I'm referring to the angle O. I know that the angle at the center it is twice the angle at the circumference. So the angle O it is twice the angle C.

[00:01:23]The angle at the center is twice the one at the circumference. Okay, something which you need to spot guys. The angle at the center and the angle at the circumference, they must be substandard by the same court. So let's go into details. Let's check into details what's happening here. If you look at the angle O, the two lines that forms the angle O, one of the line comes from A, the other line comes from B. So the angle O, it's substanded by AB.

[00:02:05]Even angle C, it is also substandarded by AB. You need to know what is to substand especially in utilian geometry.

[00:02:15]You need to know what substand the angle. In this case if I ask you what substant angle C you must be able to tell me the angle C. It is formed by these two lines.

[00:02:32]Where does these two lines come from?

[00:02:35]One comes from A.

[00:02:38]The other one comes from B.

[00:02:43]So the angle O which is the angle at the center is subended by A.

[00:02:50]Even the angle C is subended by AB. So we say O is twice angle C.

[00:02:59]You write is an angle at center is twice at circumference. We are answering uh question 3.3.1.

[00:03:17]We start by saying oh it is C. It means that the angle O it is twice X because angle C it's X. So this angle at the center here O it is 2X.

[00:03:37]Now I want you to look at this uh shape here.

[00:03:46]I want you to look at this shape.

[00:03:52]This shape here, this shape drawn with a blue color, it's a cyclic quadrilateral.

[00:04:09]We say OBTA, it's a cyclic quadriateral.

[00:04:15]If you look at all vertex of this quadrilateral OB TA, all the vertex, they are at the circumference of a circle. Which means that the opposite angles of this cyclic quad are supplementary. Which means that if I were to take angle O, add it with angle T, I must get 180°.

[00:04:46]What is the reason? Opposite angles are supplementary.

[00:04:59]when you add the opposite angles of a cyclic quad you will get 108°.

[00:05:07]So we see that angle O is 2X it means that uh 2X plus T it's 180.

[00:05:20]Then when we make t the subject of the formula t becomes 180 - 2x. If you take 2x to this side, t becomes 180 minus 2x. So you have proven 3.3.1.

[00:05:38]You have proven that t is 180 - 2x. We used two theorems here. We used the theorem that says angle at the center is twice the one at the circumference.

[00:05:51]Another theorem that we used is the theorem that has to do with a cyclic quad. We know that the opposite angles of a side quad are supplementary. So that's all you needed to solve 3.3.1.

[00:06:10]Okay. Okay.

[00:06:13]Now they are saying prove that AC it's parallel to KB. The line AC it's parallel to KB. Let us look at the line AC.

[00:06:27]The line AC it's parallel to the line KB. That's what they are saying. I must prove that AC it's parallel to KB. How do I do this? How do I prove that AC is parallel to the line KB?

[00:06:51]Uhhuh.

[00:06:53]Uhhuh. If I want to prove that AC is parallel to KB, let me think of the line that comes from C passing at B something like this. Let me call this line a transversal or it's a potential transversal.

[00:07:17]Remember a transversal grade 12s. It's a line that touches or that pass through both your parallel lines. So I want to prove that AC is parallel to KB. So I'm saying the line CBT, this straight line drawn with a blue color, it's my potential transversal. And if you think of that line as a transversal, you have somehow formed an F shape.

[00:07:48]you have formed an F shape which is looking in this manner.

[00:07:55]So I want to show that the angle C is equals to the angle B4. If I can manage to show that C, angle C is equals to angle B4, then C A will be parallel to the line KB or BK because I want to use the idea of the corresponding angles being equal. Remember corresponding angles are equal. So the angles C and B4 they form an F shape. If they are equal, CA will be parallel to KB. That's all that I want to show. I want to show that angle C is equals to B4. This C that I'm talking about, it's angle X. Already I know that C it's X. So I have to show that before also it's X. How can I show that before it's X?

[00:09:01]That's something which I need to show.

[00:09:03]How do I show that before it's X?

[00:09:13]Okay. Ahu. Ahu.

[00:09:17]Interesting guys. We are now answering 3.3.2.

[00:09:22]What I'm going to do when I'm answering this 3.3.2, 2 I'm going to say the angle K K1 is equals to the angle C. My reason is that that is the exterior angle of a cyclic quad is the angle outside the cyclic quad.

[00:09:51]Let me erase everything here and show you what I'm talking about.

[00:09:57]We don't have to rush because you have to understand these things. Remember the the main aim of this assignment is for you to learn by practicing. The idea is to practice not to memorize stuff. Don't memorize but try to understand.

[00:10:16]So this quadrilateral, this is a quadrilateral because it is it's four-sided geometric figure. This quadrilateral it's called a cyclic quad because all the vertex vertex C is at the circumference of a second. Vertex A is at the circumference of a second.

[00:10:38]Vertex K is at the circumference of a even B. So all these vertex are at the circumference making K B C A cyclic quad or you can call it A K B C or you can call it C A KB or you can call it B C A K. So this geometric figure here it's a cyclic quad. Now this angle K1 is the exterior angle of this cyclic quad. It's just outside the cyclic quad.

[00:11:15]What do we know about the exterior angle of a cyclic quad? We know that the exterior angle of a cyclic quad is equals to the opposite interior angle.

[00:11:29]The angle K1 is equals to C. That's what I was talking about. K1 is equals to C.

[00:11:39]So it means K1 it's also X. Remember C it's X. Therefore K1 it's also X.

[00:11:49]Now what am I going to do? I already know the value of angle T.

[00:11:57]I know that the angle t we said it's 180 - 2x. This angle here we set it's 180 - 2x.

[00:12:11]So what I want to do grade 12 I want to say the angle B4 plus the angle K1 plus the angle T when they are added they give us 180° these are the angles inside the triangle there is a small triangle here so I'm talking about this angles inside this triangle I'm saying when you add them, they give you 180°. Now I'm going to say B4 plus the angle K1. I said it's X plus the angle T. We have already proven on 3.3.1 that it's 180 - 2X. 180 - 2x = to 180.

[00:13:13]So I want us now to make B4 the subject of the formula.

[00:13:19]B4 equals to what are you going to do?

[00:13:24]You take this 180 to that side. So you you see that you have something like 180 minus 180.

[00:13:34]You're going to 180 - 180 it's zero. So you take this 2x to this side is going to be positive.

[00:13:42]You take even this x to that side is going to be negative.

[00:13:49]Which means that before it's equals to x.

[00:13:57]Now this is what I'm talking about guys. I am saying that for me to prove that the line CA it's parallel to the line BK.

[00:14:14]I have to show that this angle C and the angle B4 they are equal. If C is equals to B4 then C A or AC it's parallel to KB. And I've I've proven that B4 it's X and we see that C also it's X. So I'm going to say uh I'm running short of space. Let me just go and write somewhere here. I'm going to say I see that C is equals to B4.

[00:14:54]They are all equivalent to X. Now since before it's equals to angle C, I'm going to say therefore the line AC it's parallel to the line.

[00:15:13]Let me remove this. It's disting AC.

[00:15:16]It's parallel to the line KB.

[00:15:21]What is the reason? This is the converse of the theorem that says corresponding angles are equal.

[00:15:38]So please grade 12 you need to know how to prove that two lines are parallel something which is very important.

[00:15:49]Okay.

[00:16:01]So the key idea on on 3.3.2 is that when we prove that lines are parallel, we can use angles that form an F shape. Once those angles are equal then the lines will be parallel. So now let's go to triang let's prove that triangle K B KT I'm going to show that with a rich color triangle B K T it's similar.

[00:16:40]Prove that triangle B KT is similar to triangle C A T. So I'm going to say C A T.

[00:16:54]So it means is this triangle C A T. Yes. Is this triangle?

[00:17:05]It goes to T. Now guys, the the easier way of proving that two triangles are similar is by uh by showing angles that are equal. If you want to prove that two triangles are similar, you can do that by showing that some these two triangles have got angles that are equal. So, I'm going to say in this case, go let me write let me write this here cuz I really want you guys to understand.

[00:17:42]I'm going to say uh they want they want me to uh sorry, they want me to prove that the triangle B KT It's similar to the triangle C A T. So how are you going to do that? This is what you are going to do. You go to the triangle B KT. You try to check how can angle B in triangle B KT be equivalent to angle C in triangle C A T. This is what you are going to try to check first. How can B in B KT? If you look at angle B in triangle B KT, it's B4.

[00:18:36]How can B equals to the angle C in triangle C A T. And we see that C in C A T is this angle X. And we have already proven actually that the angle before it's equals to the angle C. they were this before and uh and C they are all equivalent to X. This was proven you can claim that it was proven on 3.1 2. Yes, on 3.1.2.

[00:19:15]So now you say how can I show that angle K in triangle B KT it's equivalent to angle A in triangle C A T. We see that angle K in BKT it's K1 and surely angle K1 it's equivalent to this A here which is made up of A1 and A2 and A3. Why?

[00:19:46]Because we have proven that the lines AC is parallel to the line KB. It means that this angle A and the angle K1 they are corresponding angles. they also form an F shape.

[00:20:04]Hopefully you are able to to see that.

[00:20:07]Remember we said this line it's parallel to this line. That was proven. This line is parallel to this line. And we have we have this line as a transversal.

[00:20:23]It touch both our parallel lines. So it means K1 it's equivalent to A. These are the corresponding angles. So you say K1 ah this is not that visible. You say angle K1 is equals to angle A. These are the corresponding angles.

[00:20:52]Now with the fed one actually there is no need to prove that T in triangle B KT it's equals to T in triangle C A T what you have to do the two it's enough if you have proven that B B4= to C K1 = to A you can just these ones you can just call them the fat angles of a triangle but this one I'm just going to say t I'm going to say t in the in the triangle uh bkt is equals to t in the triangle c a t. I'm going to say this is a common angle. I'm not going to say fat angle.

[00:21:44]I'm just going to say common angles.

[00:21:46]Therefore the triangle B KT it is similar to the triangle C A T. You have to conclude. You have to say therefore the triangle B KT it's similar to the triangle C A T because of angle angle angle I have proven that the angles of these two triangles actually the corresponding angles of these two triangles are equal.

[00:22:22]Now let me go to 3 4 what is is 3.3.4 3.3.4 whether they saying on 3.3.4 for they are saying if a k to kt it's 5 / 2 a k to kt is 52 a k to kt is 52 which which can also be written as A K over KT is equals to 5 over 2.

[00:23:28]Okay, they're saying if a K to KT it's 5 is 2. Determine the value of this thing is disting me. There is something here.

[00:23:42]Let me put it here this time around.

[00:23:44]Okay. So if a k to kt is 5 is to 2 determine the value of ac to kb.

[00:23:55]Okay. Normally I know that this problem of 3.3.4 it can use the solution of 3.3.3.

[00:24:08]This is typical in maths guys. When you are answering the problem of 3.4 4. What you got in 3.3 might help you when you are solving for 3.4. In 3.3 we have proven that the triangle B KT it is similar to the triangle C A T. So you have a right to do this. Pay attention. You have a right to say B K.

[00:24:40]Actually let me sorry let me write this here. So I want you to follow we have proven that triangle K I mean B KT it's similar to triangle C A T. So now I want you to say BK over you say C A.

[00:25:12]You say equals to.

[00:25:16]Now you say KT KT over 80 equals to now you say BT BT I'm sorry this one is not that visible BT t over ct.

[00:25:55]Yes. Now what you are going to do?

[00:26:00]Okay.

[00:26:02]This this bk / ca is kt over 80 bt over ct comes from the solution of 3.3. We we call this the corresponding sides of similar triangles.

[00:26:20]The corresponding sides of similar triangles are in proportion. So as long as the triangles are similar, you have a right to say their corresponding sides are in proportion. You say BK / CA equals to KT / A T equals to BT over Ct.

[00:26:40]So what you what you are going to do, you're only going to pick two from these three guys. From these three guys, you have one, two, three. You're only going to pick two. The two will help you to solve 3.4. So we want there they gave us a k over kt. So obviously we are going to pick this part.

[00:27:08]It has some at it has KT inside it.

[00:27:14]We're going to pick this part. And another thing that we are going to pick is uh BK and AC. Why? Why BK and uh BK over CA? Because it has to do with AC over KB. If you look at B K over CA, it's almost the same as AC over KB. So what I'm going to do, I'm going to say I'm using these two. Let me write here.

[00:27:46]Let me come and write here. I'm going to say B K over CA or you can decide to but you can say AC over BK.

[00:28:04]Let me call it KB. Let me say AC over KB. You can say AC over KB. I used this BK over CA. But but I wrote the denominator as the numerator, numerator as I swapped what is underneath and what is on top. So equals to.

[00:28:26]So here it means I have to say a t over kt.

[00:28:38]Yes. The reason is corresponding sides of similar triangles.

[00:28:51]Now AC over KB is what I'm looking for. I want to know how much is AC over KB. So I'm going to say AC over KB equals to I'm going to do something to this A over KT. What can I do to 80 over KT?

[00:29:17]Okay, pay attention now. grade 12 we are being told that a k / kt it's 5 / 2 from this information here from this part here we are told that a k / kt it's 5 / 2 the question is how much is a t over kt I'm going to do this I'm going to Say let a k be called 5x and kt be called 2x.

[00:30:05]Now I'm going to say 80.

[00:30:10]I have to talk about 80 because I see it here.

[00:30:15]There is 80. I'm going to say a t is made up of a k plus kt. If you look at the distance from a up until to t is made up of two part it's made up of a k added with kt.

[00:30:42]So I'm going to say now a t it's equals to a k we set it's 5x kt we said is 2x making a t 7 x.

[00:31:03]So let me try to find 80 over KT.

[00:31:09]80 over KT.

[00:31:11]Okay. Where can I write? Let me just as kids guys the way I'm writing my things is not necessarily in order. So I'm going to say let me try to find this A over KT. I'm going to say A over KT.

[00:31:32]It's equals to I set a t it's 7 x. So I'm going to say 7 x I set kt is 2x x cancel x so I have 7 / 2. So a t over kt 7 / 2. It means even AC over KB it's 7 / 2.

[00:32:04]That's it with question three. Let us go to the next question.

[00:32:13]Before we do that, let me erase this writing so that it does not distur us.

[00:32:29]Okay, I'm now going to Yes, I decided to go to question five.

[00:32:38]On question five, 5 5.1 they are saying determine the center and the radius of the circle with the equation. 5.1 is very interesting. You guys, you must be able to find the coordinates of the center of a circle from the equation of a circle which is not in the standard form.

[00:33:08]This equation of a circle is not in the standard form. This it is on 5.1 we have got x^2 + y^ 2 - 6 x + 10 y - 15 = 0. You guys must know that this is the equation of a circle but not in the standard form. The equation of a of a circle in a standard form is in this form. It's x - a 2 plus y - b² = r 2. This is the equation of a circle in a standard form. So this one x^2 + y^2 - 6 x + 10 y - 15. It is the equation of a circle but it is not in the standard form. So what we are going to do? We're going to put this equation of a circle x^2 + y^ 2 - 6 x + 10 y - 15 in the standard form. If we put it in the standard form, we will be able to see the coordinates of the center of a circle.

[00:34:29]So, we're going to say x2, we group x's together, x^2 - 6 x + y 2 + 10 y.

[00:34:46]Let's just take this -5 to this side.

[00:34:50]Equ= to 15.

[00:34:55]Now I'm going to complete the square here on x^2 - 6x. I'm going to put my focus on the number before x. The number before x, the coefficient of x is -6.

[00:35:15]I'm going to take the -6 multiply it by 1 / 2 and then square it.

[00:35:25]This is what I'm going to do. The coefficient of x multiply it by 1 / 2 and then square it. - 6 * 1 / 2 it's -3.

[00:35:39]This -3 I'm writing it because it's going to be important. It's -3 squared.

[00:35:48]Let's put this minus3 something like asteric on top. I'm saying this minus 3 we're going to need it highly.

[00:35:58]So I'm having x² - 6 x actually let me write these other ones with a red color. Let me say x^2 - 6 x I'm going to add 9. Where do I get this 9? Is this - 3^ 2 here.

[00:36:28]Okay.

[00:36:30]Then I'm going to say I'm going to say + y^ 2 + 10 y. And then now focus on the coefficient of y. The number before y is 10. I'm going to do the same thing to this 10. What am I going to do to it?

[00:36:56]I'm going to say this 10 multip and then square it.

[00:37:06]10 * 1 /2 it's 5. So it's 5 squared.

[00:37:11]This five I'm going to put something like an aesthetic next to it is going to be very important. I'm going to need it.

[00:37:19]5 squar is 25. So I'm going to add 25 and say equals to 15. Now I'm going to say what I have done on the left I must also do on the right. So if you can check on the left side I have added nine with 25. So I must do the same thing on the right side. I must add 9 with what? Again I must also add 25.

[00:38:00]So I have x2 - 6 x + 9. So you guys now need to know that this x^2 - 6 x + 9 can be factorized. You must know how to factoriize this one.

[00:38:18]So here I'm going to say just say x then you go back remember to this -3. I said you will need it. You got 9 by saying -3 squared.

[00:38:35]That's how you got 9. So you take this -3.

[00:38:41]You say x - 3 squared. That is the factor of x^2 - 6 x + 9. Actually, if you factoriize x^2 - 6 x + 9, it factor it's x - 3 * x - 3, which is the same as x - 3 2.

[00:39:01]Now you come to this one.

[00:39:05]y^ 2 + 10 y + 25. Also, it can be factorized.

[00:39:15]It's actually called a perfect square.

[00:39:18]You also factoriize by saying here you say y. In order for you to get 25, you have squared five. You set five squared.

[00:39:29]So, you add five. You say y + 5, you say squared. If you try to factoriize y^ 2 + 10 y + 25, it is y + 5 * y + 5. That's why I'm saying y + 5 2 = to 15 + 9 + 25. 15 + 25 it's 40 + 9 is 49.

[00:40:00]So you have written the equation of a circle in the standard form. It is now in the standard form.

[00:40:10]It's not difficult guys to do this. What where you need to be careful is that it's it's where you factoriize x^2 - 6 x + 9. Please make sure that you factoriize it. y^2 + 10 y + 25. Make sure that you factoriize. Now from this we can know the coordinates of a center of a circle. You can I normally take the guy inside the bracket equate it to zero or or can I just say uh the center uh no I don't want to confuse you. Let me just pick what I think it's easier.

[00:40:56]You take what is inside the bracket, you equate it to zero. You say x - 3 = to 0.

[00:41:04]Therefore, x = 3 is the xcoordinate of the center of a circle. To find the y-coordinate of a center of a circle, you take the bracket y + 5, you equate it to zero, y becomes -5.

[00:41:24]So the coordinates of a center of a circle it's 3 and -5.

[00:41:30]So that's what they wanted on uh 5.1.

[00:41:35]The coordinates of a center of a circle it's 3 and -5.

[00:41:40]What is the value of r the radius? We know that according to the formula this number at the end here this 49 represent R² it is our R² so R² it's equ= to 49 so you can use this one to find R you say R² = 49 you put the square root both sides you look for the square root of 49 the square root of 49 It's seven for sure.

[00:42:17]And the radius can never be negative. So the radius is always positive. So the radius it's positive 7. That was 5.1.

[00:42:30]So now we are going to 5.2.

[00:42:41]Let me just keep the answer here. It was three. I might need them 3 and -5. The radius is 7. Now let's go to 5.2. A second circle has the equation. Yeah, this this uh I mean this x - 2 being 2 + y + 4 being 2 = 25 is the equation of a circle in the standard form. This one.

[00:43:10]Calculate the distance between the center of the two circle or the distance between the center of the two circle.

[00:43:23]We know that the first circle has the center of 3 and -5 that is the circle of 5.1. The circle of 5.2 what are the coordinates of the center?

[00:43:36]I said to find the coordinates of a center take what is inside the bracket equate it to zero. So you can see that x here is equ= to 2 that is the center the x coordinate of a center of a circle y + 4 = 0 y = -4.

[00:44:00]So it means that the circle of 5.2 2 has its center at 2 and -4 and the question says calculate the distance between the center of this two circle. So you're going to use the distance formula to calculate the distance of the center of the two circle. So you're going to say the distance between the center it's equals to you use the distance formula you say x2 - x1 2 + y 2 - y1 squ. So the distance of the center of a circle is okay. The first circle has got the center the x the the x coordinate of a center of the first circle is three and the x coordinate of the center of the of the other circle is two. So I'm going to say 3 - 2^ 2 plus the y values of the center of a circle is -5 and -4. So I'm going to say + -5 - -4 squared.

[00:45:37]So d = to the square<unk> of 3 - 2 2 it's 1 + - 5 - - 4 - 5 is this this you just press calculator guys - 5 + 4 is -1 squar it's one so the distance between the two centers of a circle it's roo<unk> of two that's what they wanted on 5.2 the distance between the two between the two center of the circle.

[00:46:16]So 5.3 they say hence show that the circles described in question 5.1 and question 5.2 intersect each other. Okay.

[00:46:31]In order for the circle to intersect each other somehow, let's let me show you two scenarios. If the circles are uh they have got if the two circles are apart, they are far from each other. They will not intersect.

[00:46:55]Let me see the distance. Let me say uh yes d in this case I'm referring to the distance between the centers of the circle. I'm saying whenever the distance between the between the center of the two circle is greater than r1 + r2.

[00:47:23]It means that the two circles do not intersect. They they are the reason they don't intersect is that they are far from each other. The distance between their center is very huge compared to to their sum of their radius. But if the circle are close to each other, if they are close to each other, if the distance are close to each if the two circles are close to each other, the distance between the center of a circle will be lesser. It will be lesser than what? Lesser than the sum of the radius.

[00:48:11]So this means that whenever the distance is less than R1 + R2 the two circles intersect.

[00:48:21]Whenever the distance is less than R1 + R2 the two circles intersect. So please remember this guys. Whenever the distance between the two center of a circle is greater than the sum of their radius they don't intersect. So here they don't here they do intersect whenever the distance is small compared to the sum of the radius. So the distance it's already found the distance between the two center of a circle is the root of two. So I have to compare the distance to the sum of the radius.

[00:49:09]Let me find R1 and R2.

[00:49:14]What is R1 + R2?

[00:49:18]Okay.

[00:49:23]The circle of 5.1 has got the radius of 7.

[00:49:31]The circle on 5.2 has the radius of 5. I said r² = 25.

[00:49:40]r = 5. Remember this number at the end here it's your r².

[00:49:47]And if you you put r² if you put 25 inside the square root you'll see that r it's five. So now if I add the radius of this two circle is going to be 12 is going to be r1 + r2.

[00:50:08]r1 it's 7 r2 it's 5. So the sum of the radius is 12.

[00:50:15]And guess what? The square root of two which is the distance between these two circle the distance is less than the sum of the radius square root of two is less than 12. So that is why we say the two circles the circle from 5.1 and the circle from 5.2 they do intersect. How do I know the distance between the center of these two circle is less than the sum of their radius.

[00:50:55]Yes guys this topic this information of question five it's very very important. You need to make sure that you understand.

[00:51:11]I erased everything. I might need some information. Okay, think about this.

[00:51:17]Let's say this be let this be the circle of 5.1 and this other one let it be the circle of 5.2.

[00:51:33]Now I'm saying whenever the distance between the center of this two circle I've shown the distance with a black ink here. Whenever the distance between the center of the two circle is less than the radius of the circle drawn with a red color being added with the radius of a circle being drawn with a blue color.

[00:52:03]Whenever the distance is less than the sum of the radius of these two circle then the two circles intersect.

[00:52:14]Okay. 5.4 they say show that the two circle intersect along the line y = x + 5.

[00:52:26]Okay. I'm going to do this. I'm going to say I have got x2 + y^ 2 - 6 x + 10 y - 15 = 0. This is the the equation of the circle of 5.1. The circle that that they talked about on 5.1. Now on 5.2 I will prefer to write this x - 2 ^2 + y + 4 2 = to 25 in in in this in the same way I have written the equation of a circle of 5.1. I'm going I I'm I'm going to move it from being in a standard form. I'm going to say I'm talking about the equation of 5.2 2 which is x - 2^ 2 + y + 4² = to 25. I'm going to say this is the same as x 2 - 4x + 4. If you expand x - 2 2, it's x^2 - 4 x + 4. Plus, if you expand y + 4 2, it's y 2 + 8 y + 16 = 25.

[00:53:59]So I can write it as x 2 + y^ 2 - 4x + 16 y and then let me bring 25 to this side.

[00:54:19]It's going to be uh 4 - 25.

[00:54:25]4 - 25 is - 21.

[00:54:29]equals to Z. So I have these two circles. I have the circle here and I have the equation of a circle here.

[00:54:39]So I want to show that they intersect along the line y = x + 5. What I'm going to do? I'm going to what can I do? I can take uh the equation of the second circle subtract the equation of the first circle. So I'm going to say x² - x^2 it's zero and then say y^ 2 - y^ 2 it's 0. and say -4x - 6x - 4x - - 6x is going to be 2x and then 16 y -10 y it's 6 y - 21 - -5 is going to be - 6 = to 0. Then let me divide through out by 2. 2x / 2 it's x.

[00:55:54]6 y / 2 is 3 y - 6 / 2 is - 6 / 2 it's -3 = 0.

[00:56:14]So x + 3 y I'm getting x + 3 y to three. Where where am I going wrong?

[00:56:23][snorts] Did I expand this second equation very well? It's x^2 - 4x + 4 y 2 + 8 y + 16 = to 25 which is x 2 + y^ 2 - 4x + 16 y 4 - 25 is -21 = 0.

[00:56:54]x^2 - x^ 2 it's 0. y^2 - y^2 is 0 - 4x - 6 x - 4x + 6x which is 2x uh 16 y - 10 y is 6 y. Then -21 - - - - - - - - - - - - - - - - - - - - -5 it's - 6 x + 3 y - 3 = 0. I'm not getting y = x + 5.

[00:57:37]What is it that I'm not doing? Well, x^2 + y^2 - 6 x + 10 y - 15 = 0. I wonder where I'm going wrong, guys. Uh, x^2 - 4 x + 4 y 2 + 8 y + 16 = 25.

[00:58:02]Oh, I'm leaving out this 16. It's actually 4 + 16 which is 20 - 25. I'm going to be left with -5, not - 21.

[00:58:18]It's going to be -5 - 5. It's it's I'm adding like terms.

[00:58:26]It's 4x I mean it's 4 + 16 is 20. When we bring 25 this side, it's going to be minus 25.

[00:58:35]20 - 25 is - 5. So it's 2x is 2x I mean it's x^2 + y^2 - 4x + 16 y - 5 = 0. So let me take equation 2 and subtract equation 1. x^2 - x^2 is 0. y^2 - y^2 is 0. - 4x - - 6 x is 2 x 16 y - 10 y it's 6 y - 5 - -5 it's - 5 + 15 which is 10 = 0. If we divide through out by by still I'm having 6 y. It was better if I was having 2 x + 2 y = 10. 2x + Why am I going uh y Why y the y values? Did I make any mistake with the y values?

[00:59:55]16 y 10 y 16 y - 10 y it's 6 y. I'm getting guys I'm getting 2x + 6 y + 10 = 0.

[01:00:14]y * 4 y + 16.

[01:00:20]I don't know where I'm going wrong guys, but when I'm taking equation two and subtract equation one, maybe it must give me something like uh 2x uh what's what's happening here? What's happening?

[01:00:44]Okay, you'll you'll comment guys and tell me where where I was going wrong.

[01:00:48]But the procedure was right. You just need what I can tell you. You can expand 5.2. Write it in something like this form here.

[01:01:02]And then you take equation two, you subtract equation one. You must get something like y = 2x + 5.

[01:01:13]I wonder what I have missed there.

[01:01:17]So let's continue.

[01:01:21]I will check later. I have to continue guys and do other problems. So here they are saying use the definition to differentiate f ofx is equals to 1 - 3x^2.

[01:01:39]Okay, this one guys you will do it using first principle. I think you can do it.

[01:01:45]Let me go to 6.2.

[01:01:48]I just want to summarize now. y = 1 +<unk> of x being squared. If we want to differentiate this, we are going to say 1 +<unk> of x * 1 +<unk> of x.

[01:02:10]Then we we multiply throughout. We say 1 * 1 it's 1. 1 * roo<unk> of x it's roo<unk> of x.

[01:02:22]Root of x * 1 it's roo<unk> of x. Root of x * roo<unk> of x it's x guys because it's roo<unk> of x².

[01:02:38]Two cancel this y root of x * roo<unk> of x is x. So y = y = 1 + 2<unk> x + x which is the same as y = 1 + 2 x to the exponent 1 / 2 + x.

[01:03:03]So when we derive this y with respect to x, let me let me write the answer here.

[01:03:13]Here when we derive this y with respect to x, the derivative of one, it does not have x next to it. So the derivative of one, it's 0 dy / dx. The derivative of 1, it's 0. one, it's a constant number.

[01:03:33]It derivative is zero.

[01:03:36]Plus, when you derive 2x to the exponent 1 / 2, this is what you are going to do. You are going to take the exponent 1 / 2, multiply it with the number before x. 1 / 2 * 2, it's 1. So you're going to say one.

[01:04:00]You keep your variable x. You keep your variable x. The last thing that you do, you subtract your exponent by one. So your exponent it's 1 /2. If you subtract it by 1, it becomes -1 / 2.

[01:04:19]So the derivative of 2x to the exponent 1 /2, it's actually x to the exponent -1 /2. No need to write this one here. You just write x to the exponent -1 /2. The derivative of x, it's one.

[01:04:41]So that is the answer for uh 6.2 2 dy / dx is = x to the exponent -1 / 2 + 1.

[01:04:57]Okay.

[01:04:59]Now on 6.3 6.3 it's very interesting.

[01:05:06]We want to find we want to differentiate the expression inside the brackets with respect to x. So we say gx into 4 minus never never never derive if x is still underneath. Is it x to the exponent 3 or x to the exponent 2? I don't see it very well. I'm hope I'm hoping it's x to the exponent 3. I'm going to say it is 4 x to the exponent -3. You take this x to the exponent three.

[01:05:45]You take it to the top to the numerator.

[01:05:49]If it's two, it's going to be x to the exponent -2. If it's 3, it's x to the exponent -3 minus what is it? Is it 1 / x to the exponent 4? Yes. Then it becomes x to the exponent -4. this x to the exponent 4 will come on top. It becomes x to the exponent -4.

[01:06:18]So now we no longer have x as the denominator. Now we can derive the derivative of four. It's zero. It's a constant number.

[01:06:28]Here when you derive this -4x to the exponent -3 you take the exponent -3 you multiplied with the number before x which is -4. So you say -3 * -4 is pos12 you keep your variable x.

[01:06:50]The last thing you subtract your exponent with one minus 3. when you subtract it with 1, you get -4.

[01:06:59]The derivative of -x to the exponent -4 is pos4 x to the exponent -5.

[01:07:09]That is the answer for 6.3.

[01:07:14]Now let's go to question seven.

[01:07:19]Finally is my last question. I've been talking guys. I've been here question seven.

[01:07:28]They say g of x = x - 6 x - 3 x + 2.

[01:07:34]Write down the x intercept of g. Already g is in the factored form. It's factorized. The first factor it's x - 6.

[01:07:45]The second factor it's x - 3. [snorts] The fat factor it's x + 2. So what you do? You just take all these factors you you equate them to zero.

[01:08:02]Then you will know that the x intercept one of the x intercept is six.

[01:08:10]The other x intercept it's three.

[01:08:15]The other x intercept is -2.

[01:08:21]Write down the x intercepts of g. Those are the x intercepts. x = 6, x = 3, x = to -2.

[01:08:30]Determine the turning points of g.

[01:08:36]Okay.

[01:08:38]In order for us to find the turning point of g, we need to derive g. But before we derive this function g, we must put it in the standard form. I'm let me write these factors down.

[01:08:58]So I'm going to start by multiplying x - 6 with x - 3. So I'm going to say g of x = to x * x it's x 2 x * -3 is -3 x -6 * x is -6 x -6 * -3 is + 18 write x + 2, which is the same as g of x = x² - 9 x + 18 x + 2.

[01:09:54]So what I'm going to do now, I'm going to take this x, multiply it with everything in this bracket.

[01:10:06]So, it's going to be x cub - 9 x² + 18x.

[01:10:18]Then I'm going to do the same with pos2.

[01:10:22]I'm going to multiply it with everything inside this bracket.

[01:10:30]So, it's going to be 2 * x 2 x 2 * -9 -8 x 2 * 18 is 36.

[01:10:58]x cub - 9 x^2 + 2x 2. I'm adding like terms. It's -7 x^2.

[01:11:09]18x - 18x is zero.

[01:11:13]I'm left with 36.

[01:11:17]Now if I want to find the turning point of g, I have to derive g. Now g is in the standard form. then I can be able to derive it. I have to find the first derivative of g.

[01:11:36]So I'm going to say g prime x it's 3 x cub I mean 3 x² if you derive x cub is becomes 3x² - 14x the derivative of 36 is zero. Now after finding the first derivative the g prime x you put it zero. So I'm writing here guys. You now say 0 = 3x^2 - 14x. You solve for x. You are going to do that by factorizing.

[01:12:16]Uh what are the common factor? Does 3 goes into 14? I doubt 3 6 now 12. So the common factor here it's x. It's x into 3x.

[01:12:31]-4 x = 0 or 3x -4 = 0.

[01:12:42]You solve for x here. 3x = 14 x = 14 / 3. So the graphs turns at two point. It turns where x is equals to zero and it also turns where x is = to 14 /3. Now uh what was the question saying? Determine the turning points of g. You have the x turning points. You have to find the y turning points. To find the y turning points, you always substitute in the original equation of g. You can substitute in G which is in the standard form or you can substitute in G which is in a factored form. I'm going to take G which is in the standard form. I'm going to take this one. Let me show you. G of X = X cub - 7 X 2 + 36.

[01:13:48]I'm going to first look for the y turning point when x turning point is zero. So where there is x I put zero.

[01:14:02]So the y turning point is 36. So I have the turning point 0 and 36. When the x turning point is zero, the y turning point is 36. So you will do the same with 14 over3. You're going to go back to the original function. You say g of x equals to where there is x, you put 14 over 3. You say 14 over 3 cub - 7.

[01:14:35]Your x is 14 over 3 squar + 36. You press your calculator, it will tell you that when x is 14 over 3, the y turning point it's something.

[01:14:50]Please press the calculator and find that y turning point. When the x turning point is 14 over 3.

[01:15:03]Okay. Sketch the graph of g on the diagram sheet.

[01:15:08]Yeah guys, now it's possible to sketch the graph of G because you have the turning points, you have the uh x intercepts.

[01:15:23]So you can be able to sketch the graph of g.

[01:15:31]Unfortunately, I don't have a calculator. I want to sketch my graph, but let me sketch. I think I might be forced to sketch it. My graph is going to be something like this. I'm going to have the yaxis, the xaxis.

[01:15:49]Uh the x intercept we said it is three.

[01:15:56]The other one it is six.

[01:16:03]The other one is -2.

[01:16:08]These are the x intercept.

[01:16:11]Now the y intercept it's found by substituting x with zero. So you go back to the original function you put where there is x you put zero.

[01:16:24]You look for the y intercept.

[01:16:28]So 6 * 3 it's 6 * 3 it's 18 * Yes it's 36. The y intercept is 36. It's positive 36. You are using a a graph guys. You are using a diagram sheet which is given.

[01:16:50]I will just claim that 36 is here then.

[01:16:57]Okay, we found that one of the x 10 point it is zero when the y value is uh 36. Actually the graph your graph will be something like this. Your your graph will be something like this. It turns here it comes here. It turns again when x it's something. What what was it? It was when x is 14 over 3. What is 14 over 3? 3 6 9 12. It's 4 comma something. So the graph will turn again this side of four.

[01:17:40]And then it goes like this.

[01:17:43]You have you have to show the turning points here. The x turning point is 14 over 3. I don't know the y 10 point.

[01:17:51]Remember I said you will calculate it.

[01:17:54]This is the x axis, the yaxis and you must also name your graph. What is the graph that you were sketching is the graph of g?

[01:18:09]Yes. Now they are saying for which value of x is g of x times g prime of x less than zero. Okay, here guys, I will just give the solution and and ask you to think about it. Uh, how many marks is three marks? For which value of x is gx multiplied by g prime x less than zero? Okay, for me to find g of x * g prime x which is less than zero. When g of x is positive, when g of x is positive, it means that g prime x need to be negative. And when g of x is negative, g prime x needs to be positive. When I I know that when I multiply positive with negative, that's when I get an answer which is less than zero.

[01:19:13]Okay.

[01:19:14]So, gx where is it positive and where is it g prime g prime x negative.

[01:19:26]Okay.

[01:19:31]This part where from the turning point of 36 let me use a different color. From the turning point of 36 up until to the x value of 3 the g prime x there it is less than zero. When the function when the cubic function decrease its first derivative is less than zero.

[01:20:00]So when the cubic function increase its first derivative is greater than zero.

[01:20:05]So from from the from the x10 point of zero up until to the x intercept of three g prime x there it is less than zero because that's where the cubic function is going down there is decreasing.

[01:20:24]So and and from from the turning point of zero up until to three g of [snorts] x is positive because it's above the xaxis.

[01:20:34]So it means that where between the value of of uh zero and three that's when g of x multiplied with g um multiplied with that's when g of x multiplied with the first derivative of g is less than zero between zero and three. y between 0 and 3 the the graph of g of x is above the x-axis. So g of x is positive but since the function is decreasing there it means its first derivative is less than zero. [snorts] Okay again where can be my solution positive.

[01:21:28]Oh from even from what is the value of x here it's -2 even when x is less than -2.

[01:21:41]Why? When x is less than -2 the graph g of x here this part g of x it's a negative. g of x is negative because g of x we can see that it's below the x-axis at that point from this part up until to this part g of x it's negative it's below the xaxis well the first derivative there it's positive why am I saying the first derivative is positive because the function it's increasing when the function increase the first derivative is greater than zero. So this is your solution for 7.4.

[01:22:36]Good luck. All the best.