The equation of circles which passes through the origin and cuts off equal chords of length ‘a’ from the lines y = x and y = – x are

Since $\mathrm{\angle }\mathrm{AOB}=\frac{\mathrm{\pi }}{2}\Rightarrow$centres lies on either of the axes.

Let us take the case when the centre lies on x axis and let it be B(h,0)
Given OA=a
⇒OC=BC=2a​
Now in △OBC , $O{B}^2=O{C}^2+B{C}^{2_{ }}$
$$\begin{gathered} O{B}^2={\left(\frac{a}{2}\right)}^2+{\left(\frac{a}{2}\right)}^2 \\ \Rightarrow h^2=\frac{a^2}{2} \\ \Rightarrow h=\pm \frac{a}{\sqrt{2}} \end{gathered}$$

Similarly when centre lies on y axis then k=$\pm \frac{a}{\sqrt{2}}$

Hence centres = $\left(\pm \frac{\mathrm{a}}{\sqrt{2}},0\right) \mathrm{or} \left(0,\pm \frac{\mathrm{a}}{\sqrt{2}}\right)$
radius $=\frac{\mathrm{a}}{\sqrt{2}}$

Equations of the circles are
$\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2\pm \sqrt{2}\mathrm{ax}=0\text{ (or) }\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2\pm \sqrt{2}\mathrm{ay}=0$