Question

Find the equation of circle which passes through the origin and cuts off chords of length b from lines y = x and y = -x.

Answer 1

The lines y = x and y = - x are at right angles to each other and the circle passes through origin.
Also OA = OB = b
and $\angle AOB=\pi /2$
$\sqrt{(b^2+b^2)}=b\sqrt{2}$
$\therefore r=\frac{b\sqrt{2}}{2}=\frac{b}{\sqrt{2}}=O{C}_1$
Clearlr $C_1$ is centre on x - axis with co - ordinates $(b/\sqrt{2},0)$
Similarly the other centres could be
$C_2(-b/\sqrt{2},0)$ or $C_3(b/\sqrt{2},0)$
or $C_4(0-b/\sqrt{2})$
Hence the circle are
${(x\pm \frac{b}{\sqrt{2}})}^2+y^2={\left(\frac{b}{\sqrt{2}}\right)}^2$
or $x^2+y^2\pm \sqrt{2}bx=0$
or $(x-0{)}^2+{(x\pm \frac{b}{\sqrt{2}})}^2={\left(\frac{b}{\sqrt{2}}\right)}^2$
or $x^2+y^2\pm \sqrt{2}by=0$