Question

The number of circles which passes through the origin and cut off equal chords of $\sqrt{2}$ units from the straight lines $y=x$ and $y=-x$, is

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

Answer: (C)

$4$

Let the center of one such circle be $(\alpha ,\beta )$

Then, radius $=\sqrt{{\alpha }^2+{\beta }^2}$

$\therefore {\alpha }^2+{\beta }^2={\left(\frac{\sqrt{2}}{2}\right)}^2+{\left(\frac{\beta -\alpha }{\sqrt{2}}\right)}^2$ ...(1)

and ${\alpha }^2+{\beta }^2={\left(\frac{\sqrt{2}}{2}\right)}^2+{\left(\frac{\beta +\alpha }{\sqrt{2}}\right)}^2$

$\because {OC}^2={CQ}^2+{OQ}^2$ and ${OC}^2={CP}^2+{OP}^2$

where $OP=OQ=\frac{\sqrt{2}}{2}$ and $CQ=\frac{\beta -\alpha }{\sqrt{2}},CP=\frac{\beta +\alpha }{\sqrt{2}}$

Solving (1) and (2), we get

$(\alpha ,\beta )=(0,1),(0,-1),(1,0)$ and $(-1,0)$

$\therefore$ There are $4$ such circles.