Question

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

Answer 1

Let the required equation of the circle be $x^2+y^2+2gx+2fy+c=0....\left(i\right)$

Clearly, the circle passes through the points $O(0,0),A(-2,0)$ and $B(0,3).$

Putting $x=0$ and $y=0$ in (i), we get $c=0.$

Thus, (i) becomes

$x^2+y^2+2gx+2fy=0....\left(ii\right)$

Putting $x=-2$ and $y=0$ in (ii), we get $4g=4\iff g=1.$

Putting $x=0$ and $y=3$ in (ii), we get $6f=-9\iff f=\frac{-3}{2}$.

Putting $g=1$ and $f=\frac{-3}{2}$ in (ii), we get $x^2+y^2+2x-3y=0,$

which is the required equation of the circle.