 So, let's have some questions from my side, let's solve this few questions. First question is, find the locus of the pole of the line LX plus MY plus N equal to 0 with respect to the circle which touches Y axis at the origin. With respect to the circles which touch the Y axis at the origin, as you know the coordinates of the pole will change and it will trace a path, so you need to tell me the equation of that path. So, find the locus of the pole of the line LX plus MY plus N, remember LM and our constants for all the circles which touch the Y axis at the origin, which touch the Y axis at the origin. Or should I give it as a proof that question, proof that the locus will be, proof that the locus equation would be Y times LX minus N is equal to MX square. If you're done, please type done on the chat box so that I can start the discussion with this. Done? Okay. So, let's say the center of the circle which touches the Y axis is at A comma 0 and of course we know that the radius itself will be mod of A. So, we can write the equation of this circle as X minus A square plus Y square is equal to A square. In other words, it becomes X square plus Y square minus 2AX equal to 0. Now, if I have to write the, let's say the, let the pole be, let the pole be X H comma K. Because I have to find the locus of the pole, I will call it as H comma K. Okay. Let the pole be H comma K. So, if I write the pole for this circle, if I write the pole for this circle, okay, which touches the Y axis, okay, it will be X H plus Y K minus AX plus H, right? If you simplify this, it will be H minus A plus Y K minus A H equal to 0. And now we are claiming that this equation is same as LX plus MY plus N equal to 0. So, these two are the same equations, right? So, if that is the case, if they are the same equation, we can compare the coefficients. So, H minus A by L is equal to K by M is equal to minus A H by N. Remember, I have to eliminate H, sorry, A somehow because A is something which I have introduced in my equation, okay. So, from the first equation, from this equation, I can say H is equal to, sorry, A is equal to, my bad, A is equal to H minus L K by M, H minus L K by M, right? And from the second equation, I can say A is equal to minus N K by M H, minus N K by M H, right? So, both these represent the same term, so I can equate them. So, H minus L K by M is equal to minus N K by M H, okay. Let's try to simplify this, M H minus L K is equal to minus N K by H which is nothing but M H square minus L H X plus N K equal to 0, H K, I am sorry, yeah, this is K, okay. So, that becomes M H square is equal to L H minus N K and now you generalize it, now you generalize it by replacing your H with X and K with Y. So, this gives you the equation of the locus of the pole and I am sure it matches with the expression I gave you, yeah, Y times L X minus N is equal to M X square, so that's is the correct answer. Is that fine, guys? Is it clear? Any question? Please type CLR if it is clear to you so that I can move on to the next concept, okay. Great. Now, we will spend some time on the concept of common tangents, the concept of common tangents, right. We all know very well that you can have two types of common tangents drawn to a pair of circles, one is called the direct tangents, okay. And the other one is called the transverse tangents. So, there is nothing new in this concept because you are already aware of it and you have done it multiple number of times in your junior classes as well. So, what we will do is, we will directly start this concept with the concept of, we will directly start with problem solving on this concept. So let's say the first question that comes your way is, find, find all the common tangents to the circles, to the circles X square plus Y square minus 2X minus 6Y plus 9 equal to 0 and X square plus Y square plus 6X minus 2Y plus 1 equal to 0. Remember if these two circles are far apart from each other, that means they don't even touch, then you will get four answers for it, two direct tangents and two transverse tangents, okay. If at all you feel, how do you get the equation of the polar, okay. In the equation of the polar in the previous question, it is the same concept T equal to 0. So, you just please send to this, T equal to 0 is the equation of the polar that we have already learned. So, T equal to 0 means, what is T? T is X X 1, Y Y 1, G X plus X 1, F Y plus Y 1, Y plus Y 1 plus C equal to 0, right. So, wherever there was an X square, I replaced it with, this I replaced it with X X 1, this I replaced with Y Y 1, X I replaced it with X plus X 1 by 2, are you getting it? So, remember this T equal to 0, irrespective of whatever is the circle, you can always follow this formula. Is that fine, Sahyuzia? Guys, couple of things which I would like to remind you over here is that, I don't know whether you are aware of it from the junior classes or not, but whenever you are drawing circles, two circles, okay, let's say this is one circle and this is another circle, okay. When you are drawing transfers and direct tangents, so let's say I draw direct tangents to it, okay, like this, and you draw, first of all, you connect the center of these two circles, you'd realize that they'll meet this point of intersection over here, okay. So, the center of the two circles, the line drawn, the blue line will meet the direct common tangents at this point, let's say I call this point as a point, let's say this is C 1, this is C 2, let's say this is P Q R, okay, so this is the P Q R point, okay. Now, very important thing that you need to know over here is that, let's say this is radius R 1, sorry R 2, R 1, let's say, okay, then you should be all aware that R Q is to R P will be equal to R 2 by R 1, okay, so R Q is to R P would be in the same ratio as R 2 is to R 1, right, everybody is aware of it, I'm sure, so I just thought I would quickly recap this so that people who are not aware would now be aware of it, and let's say we draw the transverse tangents, let's say we draw the transverse tangents, and let's say this is the point of contact, this is 90 degree, this is the point of contact, this is 90 degree, okay, and let's say they meet at the point T over here, okay, so let's say the transverse tangents meet the line joining the centers of the circle at T, so please be informed, okay, let's say I call this point as point M, this is your M, this is your N, then this is your N, then M T is to, M T is to T N, T N will be again R 1 is to R 2, will again be R 1 is to R 2, and so will be C 1 T to C 2 T, that again will be R 1 is to R 2, so you know these two triangles are similar in this case, right, so in this case triangle C 1 M T will be similar to C 2 N T, so I'll write it down, triangles C 1 M T will be similar to triangle C 2 N T, okay, and in the case of direct common tangents triangle, triangles R P C 1, R P C 1 will be similar to triangle R Q C 2, R Q C 2, so you can write various type of relationship between them, for example you can also write R C 2 by R C 1 is again R 1 is to R 2, is again R 2 is to R 1, okay, now in case if it helps, we'll be using it in the required question as well, if done please type in the equation, first tell me the transfer tangents, that's easy to find out, so let me again write down the circles over here, so we have two circles C 1, which is x square plus y square minus 2x minus 6y plus 9 equal to 0, okay, and we have another circle C 2, okay, if you look at C 1, C 1 can be written as x minus 1 square plus y minus 3 square, correct, so I can see x square minus 2x term and y square 6y term, so I have decided it to write like this, and extra will have 9 and plus 1, so it will have a 1 over here, okay, so this is a circle which is centered at 1, 3 and is having a radius of 1, correct, so let us quickly draw a circle like that, so I'm quickly sketching a circle with center at, center at 1, 3, so this is 1, 3 and its radius is going to be 1, okay, so r is going to be 1, okay, let's look into the other circle, the other circle I can write x plus 3 the whole square, y minus 1 the whole square and additional term that I would need will be 3 square, okay, so it is a circle with center at minus 3 comma 1, so again let's draw a circle minus 3 comma 1, okay, before that I want to first figure out whether they are intersecting each other or not, okay, so what is the distance between C 1 and C 2 centers, C 1 plus C 2, this has a center at, let me call it center as 1 comma 3 and this has a center at minus 3 comma 1, so the distance between will be 1 plus 3 the whole square and 3 minus 1 the whole square, which is I think going to be 20 plus 4 root 20, root 20 is 2 root 5, okay and what is r1 plus r2, r1 plus r2 is 1 plus 3 which is equal to 4, okay, so which is greater, 4 or 2 root 5, so of course 2 root 5 will be greater, correct, so C 1, C 2 would be greater than r1 plus r2, so it's fine, they will not touch each other, so I can draw the figure like this, so you have to be very very clear with the figures guys, okay, now let's say I am trying to draw the transverse common tangents to these two, so first let us connect their centers, let us connect their centers and now I am drawing transverse common tangents to these two, like this, okay, anyways let's do only one of them, let's not crowd this diagram, let's not draw too many things which are not required, okay, now you know that this point t will divide C 1 and C 2 in the same ratio as their, as their radius, isn't it, so we know C 1 t is 2, C 2 t will be in the same ratio as their radius which is actually 3 is 2 1, right, so t divides internally the join of C 1 and C 2 in the ratio of 3 is 2 1, right, so can I find the coordinates of t, right, very easily I can find the coordinates of t, so you can find the coordinates of t as, so this is 3 is 2 1, so 3 into 1 plus 1 into 1 by 3 plus 1, correct, hope I have not written any point as incorrectly, sorry minus 3, minus 3 into 1, 1 into 3, yeah, so 3 into 1 and 1 into minus 3 by 3 plus 1 comma, comma 3 into 3 plus 1 into 1 by 3 plus 1, right, so I think t will have x coordinate as 0 and y coordinate as 10 by 4, 10 by 4 is 5 by 2, okay, now if I have to find a tangent through t, so I will say let, let the tangent drawn through, drawn through t be y minus 5 by 2 is equal to mx minus 0, okay, in other words y is equal to mx plus 5 by 2, now if this is a tangent can I say the perpendicular distance from the center will be equal to the radius, right, so I will use the distance formula, right, so I will find the distance of this point, I will find the distance between this line and this point, right, so the distance between the given line which I have assumed and the center should be equal to 3, right, so let's put it into the equation, so 1 plus 3m minus 5 by 2 mod by under root of 1 plus m square is going to be 3, isn't it, okay, so take it on the right side, square both the sides, so we will have 3m minus 3 by 2, 3m minus 3 by 2 square is equal to 9 1 plus m square, so I can always cancel off a factor of 9 from both the sides which will give me this expression, okay, yes or no, correct, now guys this would actually give you something like this, m square minus m plus 1 by 4 is equal to 1 plus m square, correct, now if you collect your m square you will get 0m square minus m minus 3 by 4 equal to 0, correct, which clearly implies that one value of m would be minus 3 by 4, so this is a quadratic, right, so I should get two values of m, so 1 is 3 by 4, let's say I call it m1 and the other two will be infinity and the other two will be infinity, the reason being I am not getting, my m squares are getting cancelled, right, so I very strategically wrote it as 0m square, right, so when I wrote it as 0m square one of the roots automatically goes to infinity and the other root is minus 3 by 4 because I have to get two answers from this, right, I have to get two tangents from this, so one of the tangent is having a slope of infinity, that's why that has not been shown in this given quadratic, the quadratic is getting converted to a linear there, but this 0 over here which I have put, right, helps you to, helps you to justify this slope which is actually infinity also, right, so we can have two equations of tangents drawn through t, 1 being y is equal to minus 3 by 4x plus 5 by 2, this is one of your answers, okay, and the other answer is going to be a line having a slope of infinity and passing through 0, so that is going to be your x equal to 0, that's going to be your x equal to 0 line, yes or no, yes or no, so these are the answers to the transverse tangents, so these are your transverse tangents, is the idea clear, this is very important because many people would not account for this, so you can have a slope of infinity of a line, so infinity is an accepted answer for m, is that fine guys, now I want you to tell me what would be the direct tangents or direct common tangents to this circle, so in case of direct common tangents guys you will be drawing these two tangents and I need this point first, so this point is basically extension of this only, so if you extend the centers it will meet at a point, let's say I call this as direct for D for direct, okay, what is the coordinates of D, if you can tell me that my 90 percent work will be done, so please type in your screen on your chat box, what is the coordinates of D, coordinates of D guys first, Aryan, Aapmesh, Arun, Siddha, Tapas, Matli, Shares, guys wake up, tell me the coordinates of D, so think as if C1, C2 are these two points, D is here, okay, and this ratio is 3 is to 1, 3 is to 1, so minus 3 comma 1 and this is 1 comma 3, so it's a case of external division, right, so tell me the formula for external division, you can take the ratio to be 3 is to minus 1 because it's a case of external division, so 3 is to minus 1, so answer will be 3 into 1, 3 into 1 and minus 1 into minus 3 which is equal to 3 again by 3 minus 1 comma, then we can write 3 into 3 which is 9 and minus 1 into minus 1 which is minus 1 into 1 which is minus 1 by 3 minus 1, yeah, so it's going to be 6 by 2 comma 8 by 2 which is going to be 3 comma 4, so coordinates of D is 3 comma 4, guys very saddening to see that you people are not able to find the point of external division point, okay, once you've got this point 3 comma 4, you can write the equation of a tangent, let the direct common tangent be y minus 4 equal to mx minus 3, okay, and then again you can use the fact that it is tangent or to any one of the circle, let's say I take the tangent from 1 comma 3, so distance of this line from 1 comma 3 should be 1 unit, okay, so let's write it as y equal to mx plus 4 minus 3m, so distance from 1 comma 3 should be equal to 1 unit, so when you write the distance you will get the expression as 3 minus m plus 3m minus 4 mod by under root of 1 plus m square is equal to 1, okay, which gives you 2m minus 1 square is equal to 1 plus m square, so 4m square minus 4m plus 1 is equal to 1 plus m square, 1 1 cancels off, so 3m square is equal to 3m square is equal to 4m, so m could be 0 or m could be 4 by 3, so putting it back over here, so two equations of the direct common tangents will be y equal to 4 when you put m as 0 and when you put it as 4 by 3 it becomes 4 by 3x plus 4 minus 4 which is 0, so it becomes y is equal to 4x by 3, so this is your second direct common tangent drawn to the circle, guys please type CLR on the screen if it is clear to you, I don't think so it is clear, you are not able to get a single answer correct guys, pretty disappointing performance given that your je is just one and a half months away, this is not how you should be behaving, you are unable to find the external point, it appears that I am writing je for you.