 Hello and welcome to the session. In this session first we are going to discuss geometrical conditions required to determine a circle. Circle has two forms of equation. One is the standard form and the other is the general form. And in each form the equation of a circle has three independent arbitrary constants. We obtain the equation of a circle. We know the values of these three constants. As there are three variables so we need three equations or conditions about the circle to find the value of these constants. Let us discuss a few geometrical conditions. First is three points are given through which the circle passes. Second is two points are given in the circle, a line on which its center lies and third is central line, point and radius etc are given. With each of these conditions three equations can be formed. Let us now discuss how to find the equation of the circle using the given condition. Let the required equation of the circle be x square plus y square plus 2gf plus 2fy plus c is equal to 0. Let us to establish three equations in g,f and c using the three given conditions like if we are given the three points 1A2, B1B2, C1C2 then put these points one by one in the equation of the circle then we shall get three of the right equations in g,f and c. Next we have if we are given the two points and a line on which its center lies say x plus y plus d is equal to 0 where d is any constant then we put two points 1 by 1 in the equation of the circle with this we get two equations with three variables g,f and c. Now at the center of the circle lies in the line plus y plus d is equal to 0 therefore the center that is gf will lie on it therefore we shall have one more equation in the form of g and f by having three of the right equations in three variables we can find the values next is with center line and radius etc. given we can find three of the right equations in g,f and c and by solving these three equations we can have the values of g,f and c then substitute the values of g,f and c then we will have the required equation of the circle let us now discuss the intersection of a line and a circle the points of intersection of a line and a circle can be obtained by solving the equations of the line and the circle let us discuss how to solve the equations of a line and a circle let the line given by x minus 2y plus 1 is equal to 0 cuts the circle x squared plus y squared is equal to 10 now we have to find the points of intersection and we take the following steps first we eliminate y or x whichever is simpler in this we find the value of x in terms of y from the equation of the line x minus 2y plus 1 is equal to 0 which implies that x is equal to 2y minus 1 now we substitute this value of x in the equation of the circle the equation of the circle is given by x squared plus y squared is equal to 10 now we put the value of x is equal to 2y minus 1 in this equation and we get 2y minus 1 the whole square plus y square is equal to 10 which can be written as 4y squared plus 1 minus 4y plus y squared is equal to 10 that is 4y squared plus y squared minus 4y plus 1 minus 10 is equal to 0 and therefore we get 5y squared minus 4y minus 9 is equal to 0 and thus we get the quadratic equation in y now let us solve this quadratic equation in y we will have the values of y the equation 5y squared minus 4y minus 9 is equal to 0 which can be written as 5y squared minus 9y plus 5y minus 9 is equal to 0 now taking y common from the first 2 terms we get y into 5y minus 9 plus taking 1 common from the last 2 terms we get 1 into 5y minus 9 is equal to 0 and therefore we get y plus 1 into 5y minus 9 is equal to 0 to have y plus 1 is equal to 0 it implies that y is equal to minus 1 and if we have 5y minus 9 is equal to 0 it implies that 5y is equal to 9 that is y is equal to 9 by 5 therefore y plus 1 into 5y minus 9 is equal to 0 implies that y is equal to minus 1 and y is equal to 9 upon 5 now the fourth step is let us substitute these 2 values of y to y minus 9 in the equation of line to have the 2 corresponding values of x therefore for y is equal to minus 1 we have x which is equal to 2y minus 1 can be written as 2 into minus 1 minus 1 that is 2 into minus minus minus 2 minus 1 which is equal to minus 3 and for y is equal to minus 5 x is equal to 2 into minus 5 minus 1 that is 2 into 9 that is 18 upon 5 minus 1 which is equal to 18 minus 5 by 5 that is 14 upon 5 we now have 2 pairs of values that is minus 3 minus 1 14 by 5 9 by 5 for the points of intersection of the line x minus 2y plus 1 with the circle x square plus y square is equal to 10 we can also determine the length of the intercept formed by the points minus 3 minus 1 and 13 by 5 9 by 5 I have line distance formula the length of the intercept is equal to square root of minus 3 minus 13 by 5 the whole square plus minus 1 minus 9 by 5 the whole square which is equal to minus 1 minus 1 square root of minus 15 minus 13 by 5 the whole square plus minus 5 minus 9 by 5 the whole square which is equal to square root of minus 15 minus 13 is minus 28 by 5 the whole square plus minus 5 minus 9 is minus 14 by 5 the whole square plus minus 5 minus So this is equal to square root of minus 28 square is 784 by pi square that is 25 plus minus 14 square that is 196 upon pi square that is 25. So we get square root of 980 by 25 which is equal to 196 by pi and therefore we get square root of 196 by pi is equal to 14 upon square root of pi which can also be written as 14 into square root of 5 by 5 therefore length of the intercept is given by 14 into square root of 5 by 5 units. This completes our question for you enjoyed this question.