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

$\frac{1}{\sqrt{2}}$

Explanation for the correct option.

Step 1: Find the equation of the common tangent

Any tangent to ${\mathrm{y}}^2=4\mathrm{x}$ is $\mathrm{y}=\mathrm{mx}+\frac{1}{\mathrm{m}}$, tangent to ${\mathrm{x}}^2=4\mathrm{y}$ with slope $\mathrm{m}$ is $x=\left(\frac{1}{m}\right)y+m$ or $y=mx-m^2$.

$\begin{array}{rcl}\frac{1}{\mathrm{m}}& =& -{\mathrm{m}}^2\\ {\mathrm{m}}^3& =& -1\\ \mathrm{m}& =& -1\end{array}$
Substituting the value of $m$ in equation $\mathrm{y}=\mathrm{mx}+\frac{1}{\mathrm{m}}$we get the equation of common tangent as:

$x+y=-1$

Step 2: Find value of $c$

As the line $x+y=-1$ touches the circle $x^2+y^2=c^2$ it means that the radius is equal to the distance between the center & tangent.

$\begin{array}{rcl}& \Rightarrow & \left|c\right|=-\frac{1}{\sqrt{2}}\\ & \Rightarrow & c=\pm \frac{1}{\sqrt{2}}\end{array}$

Hence, the $c$ is equal to $\pm \frac{1}{\sqrt{2}}.$