 Hello and welcome to the session. In this session we are going to discuss interest on the axis in case of a circle. As we know that if a point lies on x-axis in the coordinate system then its y-coordinate is 0. That is y is equal to 0. Similarly on y-axis the x-coordinate is 0. That is x is equal to 0. Now let us consider the general equation of the circle x-square plus y-square plus 2 gx plus 2 fy plus c is equal to 0. To find x-intercept we put y is equal to 0 in the equation of the circle. Therefore we have x-square plus 0-square plus 2 gx plus 2 f into 0 plus c is equal to 0. And this will result into a quadratic equation in x only that is x-square plus 2 gx plus c is equal to 0 and on solving this equation that is x-square plus 2 gx plus c is equal to 0 We will get two values of x the algebraic difference of the two values of x gives the length of the x-intercept. Now to find y-intercept we put the value of x is equal to 0 in the equation of the circle. Therefore we get 0-square plus y-square plus 2 g into 0 plus 2 fy plus c is equal to 0 and this results into a quadratic equation in y only that is y-square plus 2 fy plus c is equal to 0. Now on solving this equation that is y-square plus 2 fy plus c is equal to 0 we will get two values of y the algebraic difference of the two values of y gives the length of the y-intercept. Let us take an example find the length of the intercepts on the circle x-square plus y-square plus 2x plus 2y minus 3 is equal to 0 for x-intercept we put y is equal to 0 in the equation of the circle. Therefore the equation will reduce to x-square plus 2x minus 3 is equal to 0 after putting y is equal to 0 in the given equation which can also be written as x-square plus 3x minus x minus 3 is equal to 0. On taking x common from the first two terms we have x into x plus 3 and taking minus 1 common from the last two terms we have minus 1 into x plus 3 which is equal to 0 which implies that x minus 1 into x plus 3 is equal to 0. Therefore x is equal to 1 and minus 3 x minus 1 is equal to 0 implies that x is equal to 1 and x plus 3 is equal to 0 implies that x is equal to minus 3. Hence the circle cuts off an interface of three units in the negative direction and one unit in the positive direction. Therefore length of the x-intercept which is equal to the algebraic difference of the two values of x that is minus 3 minus 1 which is equal to minus 4 therefore length of the x-intercept will be four units. Now to find out the y-intercept we put x is equal to 0 in the given equation of the circle and we get y-square plus 2y minus 3 is equal to 0 which implies that y is equal to minus 3 and 1. We have the equation y-square plus 2y minus 3 is equal to 0 which can be written as y-square plus 3y minus y minus 3 is equal to 0. Now taking y common from the first two terms we get y into y plus 3 and taking minus 1 common from the last two terms we get minus 1 into y plus 3 which is equal to 0. Therefore we get y minus 1 into y plus 3 is equal to 0 which implies that y is equal to 1 and y is equal to minus 3. The circle cuts off, intercepts off in the negative direction and one unit in the positive direction. Therefore length of the y-intercept which is equal to the algebraic difference of the two values of y that is minus 3 minus 1 which is equal to minus 4. Therefore length of the y-intercept will be 4 units. This completes our session. Hope you enjoyed this session.