 And today I'm going to start my discussion with circle. In fact, I'm quickly going to take you through main and JEE advance concepts. But at the same time, I would quickly revise whatever you have learned in circles so far without much delay. So let us start our discussion with the first conic, which is the simplest of all, which is the circle. I'm not going to take a lot of time for revision. Just a few things that I want you to all know. First, you should know the equation of the circle in various forms. So we'll start with equation in various forms. First one is the center radius form, which is quite familiar to all of you guys. So if you have a circle whose center is at h, k, and whose radius is going to be a units, its equation is given as x minus h whole square y minus k whole square equal to a square. No doubt about it. Very simple concept. Second is the equation of a circle in parametric form. For the same circle, which I have mentioned right now over here, for the same circle, if you write the equation in parametric form, we write it as x equal to h plus a cos theta and y is equal to k plus a sine theta. Here theta is a parameter. If you try to eliminate this parameter, you will end up getting the center radius form of the equation of the circle. Quickly moving on, this is all known to you. So I'm not going to spend too much time on it. Third is the general form. The general form of the equation of a circle, as we have already seen in case of a general second degree equation, a and b should be same. So what I did is I assumed them to be 1 or you say you can divide throughout with a or b. And this becomes the general form of the equation of a circle where the center of the circle is given as minus g minus f. And the radius is given as under root of g square plus f square minus c. So this is also known to you guys. No need to waste too much time on this. The fourth form is the diametrical form. Diametrical form of the equation of a circle. So whenever a circle is given to you, whose extremities of the diameter. So let's say this is the diameter. So let's say extremities of the diameter are given to us as x1, y1, and x2, y2. Then we write the equation of this circle as x minus x1, x minus x2 plus y minus y1 times y minus y2 equal to 0. So nothing hard and fast about this. This was all known to you. Fourth is the equation of a circle. Equation of a circle passing through, passing through three non-colinear points. Three non-colinear points. Very important, normally people tend to forget this. It is in the form of a determinant where the first of the determinant is x2 plus y2 plus xy1. Second row is x1, y1 square. Passing through three non-colinear points. So let the point be x1, y1, x2, y2, and x3, y3. x3, y3. So it will be a determinant given by x2 plus y2, xy1. x1 square plus y1 square, x1, y1, 1. x2 square plus y2 square, x2, y2, 1. And x3 square plus y3 square, x3, y3, 1. Now guys, how does this equation come about? This equation comes in a very simple way. So let the equation be x square plus y square plus 2g. So I'm going to prove this quickly. 2fy plus c equal to 0. Let this be the equation of the circle. And since it satisfies these three points, x1, y1. So x1, y1 will satisfy this equation. x2, y2 will also satisfy this condition. And x3, y3 will also satisfy this condition. Now, all I need to do is I need to eliminate gf and c from this equation. Now, again, treat these numbers as coefficients. Treat these numbers as coefficients of gf and c and 1. Treat them as coefficients. So these are all coefficients of g, f, and c, and 1. And the determinant formed by them should be equal to 0. So the determinant formed by all these coefficients should be 0 because your gf, c, et cetera, and 1 cannot be trivial. So treat this in light of a trivial solution of system of homogeneous equation. You directly get this expression. I would request you to remember this expression because it proves to save a lot of time. You cannot afford to sit and derive this in the examination hall. And it's very easy to use this formula. So these are the various forms of the equation of a circle. Now, we are moving to the next concept, intercepts made by the x and the y-axis. In fact, you can say intercept made on the axis is by a circle. So let's say there is a circle like this. And this is my y-axis and this is my x-axis. So basically, if I want to know how much intercept has been cut by this circle, let's say it's x square plus y square plus 2 gx plus 2 fy plus c equal to 0 on the x-axis. That means I need to know what is the length AB. Please note that AB is not the diameter. It is the intercept cut on the x-axis. So how can I find AB? Very simple. You'll say put y as 0. When you put y as 0, you get x square plus 2 gx plus c equal to 0. And we know that this point will be something like x1 comma 0. This point will be something like x2 comma 0. And what I need is mod of x1 minus x2. So how will I get that? Very simple. You know its roots are going to be x1 and x2. So x1 plus x2 will be minus 2g. x1 into x2 will be c. So we know the formula x1 minus x2 whole square is x1 plus x2 whole square minus 4x1 x2. That is going to be 4g square minus 4c. So x1 minus x2 mod will be under root of this, which is going to be 2 under root g square minus c. So remember this result. It's very, very helpful in many cases. However, you can also find it out by using the fact that you know the distance of the origin from the x-axis. That is nothing but the ordinate of the center and you know the radius. So you can always find this length and double it up. That's another way. But it's very important to understand or look at this concept from quadratic equation point of view as well. Similarly, the intercept cut on the y-axis. Let me call it as CD. Intercept cut on the y-axis will be 2 under root f square minus c. So please remember this result. CD is 2 under root f square minus c. And AB is 2 under root g square minus c. Please remember this result as well. Moving on, next concept is position of a point with respect to a circle. With respect to a circle. So if you have been given any circle, whose equation is x square plus y square plus 2 gx plus 2 fy plus c equal to 0, the first thing that I want to discuss over here, guys, this term henceforth would be referred to by the name s. s stands for second degree equation. So this is a second degree term. So I will represent this term by s. So this expression is x. This expression is called s. Now when I substitute, let's say I talk about this point, x1, y1 being inside the circle. Let's say it is at position a. When you substitute x1, y1 into in place of x and y, I would call that expression as s1. So if the point is located within the circle, that means in case where the point is at a, you would realize that your s1 would be negative. That is, the value of this term that would be obtained by substituting such a point which is within the circle would be a negative number. If it lies on the circle, that is your case b, of course, the point is going to satisfy the equation. So s1 is going to be 0. And if it is outside the circle, let's say at point c, so for case c, it would be greater than 0. By the way, s1 is called the power of the point with respect to the circle. We say power of a point, if it is negative, it the point lies within the circle. If the power of a point is equal to 0, it lies on the circle. And if power of the point is greater than 0, it lies outside the circle. And one important thing I wanted to discuss over here is that if the point actually lies outside the circle, let's say I take a small case over here. Let's say this is the point x1, y1. So when you draw a tangent from this line onto the circle, so let's say I call it as pt, then length of pt is going to be under root of s1. Please remember this. So if you square root the power of a point, you will end up getting the length of the tangent drawn from the external point onto the circle. And as you can clearly see, if the point is within, the length of the tangent will come out to be imaginary. And if the point is on the circle, it will come out to be 0. And if it is outside the circle, it will come out to be positive.