## Task 3 - 140813

Given a circle $k_{1}$ with radius $r_{1}$ and center $M$. Around $M$, a circle $k_{2}$ is to be drawn such that the area of the annulus between $k_{1}$ and $k_{2}$ is three times the area of the circle $k_{1}$.

Calculate the radius $r_{2}$ of the circle $k_{2}$!
The area $A_{1}$ of the circle $k_{1}$ is given by $A_{1}=\pi r_{1}^{2}$, and the area $A_{R}$ of the annulus between $k_{1}$ and a circle $k_{2}$ with radius $r_{2}$ and center $M$ is

$$
A_{R}=\pi\left(r_{2}^{2}-r_{1}^{2}\right)
$$

Therefore, the condition required in the problem is satisfied if and only if $3 \pi r_{1}^{2}=\pi\left(r_{2}^{2}-r_{1}^{2}\right)$. This is equivalent to $4 r_{1}^{2}=r_{2}^{2}$, and due to $r_{1}>0, r_{2}>0$, this is equivalent to $2 r_{1}=r_{2}$.

Thus, the circle $k_{2}$ has the required property if and only if its radius is exactly twice the radius of the circle $k_{1}$.