Question

The number of circles which passes through the origin and makes intercept of length $8$ units and $6$ units on the coordinate axes respectively, is

Answer 1

Let the required circle be $x^2+y^2+2gx+2fy+c=0$
$\because$ Circle passes through origin, $c=0$
Given, $2\sqrt{g^2-c}=8\Rightarrow g^2=16\Rightarrow g=\pm 4$
$2\sqrt{f^2-c}=6\Rightarrow f^2=9\Rightarrow f=\pm 3$
$\therefore$ Equation of circle : $x^2+y^2\pm 8x\pm 6y=0$
$\therefore 4$ circles are possible.