The equations of the circle which pass through the origin and makes intercepts of lengths 4 and 8 on the $x$…
- $x^2+y^2 \pm 4 x \pm 8 y=0$
- $x^2+y^2 \pm 2 x \pm 4 y=0$
- $x^2+y^2 \pm 8 x \pm 16 y=0$
- $x^2+y^2 \pm x \pm y=0$
Solution

$\therefore$ Required equation of circle is $\begin{aligned} & (x \pm 2)^2+(y \pm 4)^2=20 \\ \Rightarrow \quad & x^2+y^2 \pm 4 x \pm 8 y=0 \end{aligned}$
Asked in: AP EAMCET 2009