## Task B-4.4.

Determine all real numbers $k$ for which the circles given by the equations $(x+1)^{2}+y^{2}=9$ and $(x-k)^{2}+(y-2 k)^{2}=4$ have only one common point.
## First solution.

Let the center and radius of the first circle be $S_{1}(-1,0), r_{1}=3$, and the second circle $S_{2}(k, 2k), r_{2}=2$.

The circles will have one common point if they touch each other externally or internally. If the circles touch each other externally, then $\left|S_{1} S_{2}\right|=r_{1}+r_{2}=5$.

Then, $(k+1)^{2}+4 k^{2}=25$, or $5 k^{2}+2 k-24=0$.

The solutions to this quadratic equation are $k_{1}=2, k_{2}=-\frac{12}{5}$.

If the circles touch each other internally, then $\left|S_{1} S_{2}\right|=\left|r_{1}-r_{2}\right|=1$.

Then, $(k+1)^{2}+4 k^{2}=1$, or $5 k^{2}+2 k=0$.

The solutions to this quadratic equation are $k_{3}=0, k_{4}=-\frac{2}{5}$.

Finally, the desired numbers $k$ are $-\frac{12}{5},-\frac{2}{5}, 0$ and 2.