 Hello and welcome to the session. My name is Mansi and I'm going to help you with the following question. The question says, find the equation of the circle passing through the point 0,0 and making intercepts a and b on the coordinate axis. So let us start with the solution to this question. First of all, we see that equation of the circle with center at h k and radius r is given by x minus h the whole square plus y minus k the whole square equals to r square. Now it's given to us in the question that circle is passing through the point 0,0. So first of all what we do is we put x and y equal to 0,0. Let us call this equation a. So we put this in equation a and we get 0 minus h the whole square plus 0 minus k the whole square equals to r square or h square plus k square is equal to r square and we call this equation 1. Now it's also given to us in the question that the circle is also passing through a and b on the coordinate axis that means the circle passes through the point a,0 and 0,b. Because it's given to us that they are passing through coordinate axis making intercepts a and 0,a and b. So in the x-axis it would meet the coordinate axis at a,0 and the y-axis at 0,b. So we can say that circle passes through a,0 therefore putting x and y equal to a,0 in equation a we get a minus h the whole square plus 0 minus k the whole square is equal to r square or a square plus h square minus 2a h plus k square is equal to r square we call this equation 2. Again we put x and y to be equal to 0,b in equation a that is x minus h the whole square because y minus k the whole square equal to r square we get 0 minus h the whole square plus b minus k the whole square is equal to r square or h square plus k square plus b square minus 2b k is equal to r square we call this equation 3. Now we subtract equation 1 from equation 2 so subtract 1 from 2 we get h square plus k square plus a square minus 2a h minus h square plus k square equals to r square minus r square. Now h square gets cancelled with minus h square k square gets cancelled with minus k square and we have a square minus 2a h equal to r square gets cancelled with minus r square and we have 0. This implies a is minus 2h will be equal to 0 because either a equal to 0 or a minus 2h equal to 0 since a is not equal to 0 so we have this and this implies that h is equal to a by 2. Again subtracting equation 1 from equation 3 we get h square plus k square plus b square minus 2b k minus h square plus k square is equal to r square minus r square. Now we see h square gets cancelled with minus h square k square gets cancelled with minus k square and we have b square minus 2b k is equal to 0. This implies b minus 2k equal to 0 and this implies k is equal to b by 2. Now subtracting 1 now what we do is we simply put h equal to a by 2 and k equal to b by 2 in equation 1 to find out the value of r square. So this implies that r square is equal to a by 2 the whole square plus b by 2 the whole square. Now putting all the obtained values in this equation we get x minus a by 2 the whole square plus y minus b by 2 the whole square is equal to a by 2 the whole square plus b by 2 the whole square or we can say x square plus a by 2 the whole square plus or minus ax plus y square plus b by 2 the whole square minus b y is equal to a by 2 the whole square plus b by 2 the whole square. Now a by 2 the whole square plus b by 2 the whole square get cancelled with a by 2 the whole square plus b by 2 the whole square in the right hand side and this gives us x square plus y square minus ax minus b y equal to 0. So our answer to the question is that the required equation of the circle is x square plus y square minus ax minus b y equal to 0. So this is our answer to the question. I hope that you understood the question and enjoyed the session. Have a good day.