 Hello and welcome to the session. In this session we will derive the equation of a circle of given center and radius using Pythagorean theorem. We will also discuss method of completing square to find center and radius of circle. First of all we shall derive general equation of circle with center hk and radius r using Pythagorean theorem. This is the circle on coordinate plane with center c having coordinates hk and radius r let p with coordinates xy be any point on the circle. Now from point p draw a vertical line and from point c draw a horizontal line let intersection of these two lines be point q then we see that triangle cpq is a right angle triangle. Now see the coordinates of point q its x coordinate will be same as the x coordinate of point p because both lie on same vertical line that is its x coordinate will be x and its y coordinate will be same as y coordinate of point c because both points lie on same horizontal line. So the y coordinate of q will be k so we say that coordinates of point q are xk. Now in right angle triangle cpq vertices are c with coordinates hk p with coordinates xy and q with coordinates xk. Now let us find the length of sides here we know that cp is the radius of the circle so length of side cp will be equal to r that is the radius of the circle. Length of side cq will be equal to modulus of x minus h because horizontal distance between any two points with coordinates x1, y1 and x2, y2 on coordinate plane is given by modulus of x2 minus x1. Length of the side pq is given by modulus of y minus k because vertical distance between any two points with coordinates x1, y1 and x2, y2 on coordinate plane is given by modulus of y2 minus y1. So we have cp is equal to r cq is equal to modulus of x minus h and pq is equal to modulus of y minus k. Now using Pythagorean theorem we have cq square plus pq square is equal to cp square that is sum of the squares of the two adjacent sides of a right angle triangle is equal to the square of the hypotenuse which implies that cq square that is x minus h whole square plus pq square that is y minus k whole square is equal to cp square that is r square which is the equation of the circle with center hk and radius r. In general the center radius equation of a circle with center hk and radius r is given by x minus h whole square plus y minus k whole square is equal to r square. This is also called standard form of the equation of circle. Let us take an example. Find the equation of circle with center 2 minus 2 and radius 3. Since general equation of circle in center radius form is given by x minus h whole square plus y minus k whole square is equal to r square. So here the ordered pair hk is equal to the ordered pair 2 minus 2 and r is equal to 3. So putting values of hk and r in this equation we get x minus 2 whole square plus y minus of minus 2 whole square is equal to 3 square which implies that x minus 2 whole square plus y plus 2 whole square is equal to 3 square that is 9. So x minus 2 whole square plus y plus 2 whole square is equal to 9 is the required equation. Now we discuss method of completing square to find center and radius of a circle. The equation of a circle is not always given in standard form. So we have to use the method of completing the square to rewrite the equation into standard form and we are already familiar with the method of completing square for a quadratic equation. We will apply same method to make perfect square in the given equation of circle. Let us consider the following example. The equation of circle is given by x square minus 4x plus y square plus 6y minus 3 is equal to 0. We have to find its center and radius. So we will first convert it into standard form for that we need to make perfect square. We will follow the following steps. The first step is make coefficients of x square and y square 1 by dividing the whole equation by the common coefficient of x square and y square. Here it is already 1 so we leave it as it is. The second step is keep terms with variables x and y on one side and bring the constant term on the other side. So the equation becomes x square minus 4x plus y square plus 6y is equal to 3. In the next step we put the x variable and y variable terms in separate brackets. So we get x square minus 4x the whole plus y square plus 6y the whole is equal to 3. Now x square minus 4x is one quadratic in x and y square plus 6y is quadratic in y. So we make perfect squares in the brackets. For x variable terms we divide the coefficient of x by 2 and then square the result we obtain. Then we add this obtained number inside the bracket and also add this number to the right hand side of the equation so that we don't change the original equation. Here we have x square minus 4x so we will add the square of half the coefficient of x that is we have half the coefficient of x that is 1 by 2 into minus 4 the whole square which is equal to minus 2 whole square that is equal to 4. So we will add 4 inside the brackets and also add 4 to the right hand side of the equation. We shall follow the same method for terms in y variable. Here we have y square plus 6y and we will add square of half of the coefficient of y that is 1 by 2 into 6 whole square which is equal to 9. So we add 9 inside the brackets and we will also add 9 to the right hand side of the equation. So we get x square minus 4x plus 4 the whole plus y square plus 6y plus 9 the whole is equal to 3 plus 4 plus 9 which implies that x square minus 4x plus 4 the whole plus y square plus 6y plus 9 the whole is equal to 3 plus 4 that is 7, 7 plus 9 is 16. Now we know that a minus b whole square is equal to a square minus of 2ab plus b square and a plus b whole square is equal to a square plus 2ab plus b square. So x square minus 4x plus 4 can be written as x whole square minus 2 into 2 into x plus 2 square. Now on comparing with this equation we can write it as x minus 2 whole square. Similarly we can write y square plus 6y plus 9 as y whole square plus 2 into 3 into y plus 3 square and on comparing it with this equation we can write it as y plus 3 whole square. So we have x minus 2 whole square plus y plus 3 whole square is equal to 16. Now 16 can also be written as 4 square and this is the standard form of the equation of circle that is x minus h whole square plus y minus k whole square is equal to r square. So here h is equal to 2 and k is equal to minus 3. Thus center will be given by the ordered pair 2 minus 3 also radius is equal to 4. Thus in this session we have derived the equation of circle of given center and radius using the Pythagorean theorem and we also discussed how to use method of completing square to find center and radius of circle. This completes our session hope you enjoyed this session.