Question

Equation of the circle through origin which cuts intercepts of length a and b on axes is
(a) x² + y² + ax + by = 0
(b) x² + y² − ax − by = 0
(c) x² + y² + bx + ay = 0
(d) none of these

Answer 1

(b) x² + y² − ax − by = 0

Centre of the circle is $\left(\frac{a}{2},\frac{b}{2}\right)$ and its radius is $\sqrt{{\left(\frac{a}{2}\right)}^2+{\left(\frac{b}{2}\right)}^2}=\frac{1}{2}\sqrt{a^2+b^2}$.

Equation of circle:
${\left(x-\frac{a}{2}\right)}^2+{\left(y-\frac{b}{2}\right)}^2=\frac{1}{4}\left(a^2+b^2\right)$
$\Rightarrow$${\left(2x-a\right)}^2+{\left(2y-b\right)}^2=\left(a^2+b^2\right)$
$\Rightarrow$$4x^2+a^2-4ax+4y^2+b^2-4by=a^2+b^2$
$\Rightarrow$$x^2-ax+y^2-by=0$