Question

A straight line $x=y+2$ touches the circle $4(x^2+y^2)=r^2$. The value of $r$ is:

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

Answer 1

The correct option is B $2\sqrt{2}$

We have equation of the circle is $x^2+y^2=\frac{r^2}{4}$

or, $x^2+y^2={\left(\frac{r}{2}\right)}^2$

$x=y+2$ is tangent to the circle.

Centre of the circle: $(0,0)$

Tangent is always perpendicular to the radius of the circle.

So, $\frac{r}{2}=\frac{|x-y-2|}{\sqrt{2}}$ at $(x=0,y=0)$

or, $r=\frac{4}{\sqrt{2}}=2\sqrt{2}$

Hence, B is the correct option.