Question

A circle $C_1$ of radius $2$ touches both x-axis and y-axis. Another circle $C_2$ whose radius is greater than $2$ touches circle $C_1$ and both the axes. Then the radius of circle $C_2$ is

• A. $6-4\sqrt{2}$
• B. $6+4\sqrt{2}$
• C. $6-4\sqrt{3}$
• D. $6+4\sqrt{3}$

Answer 1

The correct option is B $6+4\sqrt{2}$

The first circle touches both axes and radius is $2$ units.

Hence center of the circle is $(2,2)$

Let the radius of other circle be $a$ and this circle also touches both the axes.

Hence center of circle is $(a,a)$

This circle touches the first circle

Hence $\sqrt{{(a-2)}^2+{(a-2)}^2}=a+2$

On squaring both the sides, we get

$a^2-12a+4=0\Rightarrow a=\frac{12\pm \sqrt{{\left(12\right)}^2-4\times 4\times 1}}{2}$

$=\frac{12\pm \sqrt{128}}{2}=6\pm 4\sqrt{2}$

But $a>2$, therefore $a=6-4\sqrt{2}$ is neglected,

Hence, $a=6+4\sqrt{2}$.