Question

Find the equation of a circle passing through the origin and intercepting lengths $a$ and $b$ on the axes.

Answer 1

General equation of circle $=x^2+y^2+2gx+2fy+c=0.....\left(1\right)$

Given that the points $(0,0),(a,0)$ and $(0,b)$ lie on the circle.

Therefore, all these points will satisfy the above equation

$0+0+0+0+c=0$

$\Rightarrow c=0$

Also,

$a^2+0+2ag+0+0=0$

$\Rightarrow 2g=-a$

Also,

$0+b^2+0+2bf+0=0$

$\Rightarrow 2f=b$

Substituting all these values in equation $\left(1\right)$, we have

$x^2+y^2-ax-by=0$

Hence the equation of circle is $x^2+y^2-ax-by=0$.