Question

Find the equation to the circles which pass through the origin and cut off equal chords $a$ from the straight lines $y=x$ and $y=-x$.

• A. $x^2\pm \sqrt{2}ay+y^2=0$
• B. ${-x}^2\pm \sqrt{2}ay+y^2=0$
• C. $x^2\pm \sqrt{2}ay-y^2=0$
• D. None of the above

Answer 1

The correct option is A $x^2\pm \sqrt{2}ay+y^2=0$

From the figure, it is clear that the centre of the circle will lie on either $x$ or $y$ axis.

Let us take the case when the centre lies on $x$ axis and let it be $B(h,0)$

Given $OA=a$

$\Rightarrow OC=BC=\frac{a}{2}$

Now in $△OBC$ $,$ ${OB}^2={OC}^2+{BC}^2$

$\Rightarrow {OB}^2={\left(\frac{a}{2}\right)}^2+{\left(\frac{a}{2}\right)}^2$

$\Rightarrow h^2=\frac{a^2}{2}$

$\Rightarrow h=\pm \frac{a}{\sqrt{2}}$

So the centre of the circle is $(\pm \frac{a}{\sqrt{2}},0)$ and radius is $\frac{a}{\sqrt{2}}$ and its equation is

${(x\pm \frac{a}{\sqrt{2}})}^2+{(y-0)}^2={\left(\frac{a}{\sqrt{2}}\right)}^2$

$x^2+\frac{a^2}{2}\pm \sqrt{2}ax+y^2=\frac{a^2}{2}$

$x^2\pm \sqrt{2}ax+y^2=0$

Similarly when centre lies on $y$ axis centre will be $(0,\pm \frac{a}{\sqrt{2}})$ and radius $\frac{a}{\sqrt{2}}$ and its equation will be

$x^2\pm \sqrt{2}ay+y^2=0$