Q11 of 27 Page 11

Find the equation of the circle passing through the origin and cuts off intercepts -2 and 3 from the x-axis and y-axis respectively.

Let the required equation of circle be


x2 + y2 + 2gx + 2fy + c = 0 ….. (1)


As circle passes through (0,0),(-2,0)(0,3)


Putting (0,0) in (1) to get,


0+0+0+0+c = 0


⇒ c= 0


Thus (1) becomes,


x2 + y2 + 2gx + 2fy = 0 ….. (2)


Putting (-2,0) in (2) to get,


(-2)2 + 02 – 4g + 0 = 0


⇒ 4g = 4


g = 1


Putting (0,3) in (2) to get,


0 + 32 + 0 + 6f = 0


9 + 6f = 0


6f = -9


f = -3/2


Putting g = 1, f=-3/2 in (2) we get,


x2 + y2 + 2x – 3y= 0


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