Question

If the common tangent to the parabolas, $y^2=4x$ and $x^2=4y$ also touches the circle, $x^2+y^2=c^2$, then $c$ is equal to:

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{1}{2\sqrt{2}}$

Answer 1

The correct option is C $\frac{1}{\sqrt{2}}$

$y=mx+\frac{1}{m}$
$\Rightarrow x^2=4(mx+\frac{1}{m})$
$\Rightarrow x^2-4mx-\frac{4}{m}=0$
$D=0$
$\Rightarrow 16m^2+\frac{16}{m}=0$
$\Rightarrow 16\left(\frac{m^3+1}{m}\right)=0$
$\Rightarrow m=-1$
$\Rightarrow y=-x-1$
$\Rightarrow x+y+1=0$
Perpendicular distance of the tangent from the centre of circle is equal to radius.
$\therefore p=c$
Now, $\left|\frac{0+0-1}{\sqrt{2}}\right|=\left|c\right|$
$\Rightarrow c=\pm \frac{1}{\sqrt{2}}$