Given are the externally tangent circles $k_{1}, k_{2}$. What will be the radius of the circle that is internally tangent to both $k_{1}$ and $k_{2}$, and whose radii at the points of tangency are perpendicular to each other? Also, construct this circle!
Let the radii of $k_{1}, k_{2}$ be $r_{1}, r_{2}$ respectively, their centers be $O_{1}, O_{2}$, the center of the tangent circle $k$ be $O$, its radius be $R$, and the tangency point of $k$ and $k_{1}$ be $E_{1}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_0da87cda70deaa0e0d8cg-1.jpg?height=802&width=1287&top_left_y=279&top_left_x=408)

The ray $E_{1} O_{1}$ contains $O$ and $R=E_{1} O>E_{1} O_{1}$, since $k_{1}$ is internally tangent to $k$. Similarly, $O$ is on the ray $E_{2} O_{2}$, where $E_{2}$ is the tangency point of $k$ and $k_{2}$. Furthermore, we know that the segment $O_{1} O_{2}$ contains the tangency point $E$ of the circles $k_{1}$ and $k_{2}$. According to the conditions, the triangle $O_{1} O O_{2}$ is a right triangle, and from the above, we have

$$
O_{1} O_{2}=r_{1}+r_{2}, \quad O_{1} O=R-r_{1}, \quad O_{2} O=R-r_{2}
$$

By the Pythagorean theorem, we have

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

from which

$$
R=\frac{r_{1}+r_{2}+\sqrt{\left(r_{1}+r_{2}\right)^{2}+4 r_{1} r_{2}}}{2}
$$

Only the positive sign is possible, since $r_{1}+r_{2}-\sqrt{\left(r_{1}+r_{2}\right)^{2}+4 r_{1} r_{2}}<0$. Therefore, if the circle exists, its radius can only be this value. If $r_{1}=r_{2}$, then the tangent circle can be easily constructed, since in this case $O O_{1} O_{2}$ is an isosceles right triangle with hypotenuse $O_{1} O_{2}=r_{1}+r_{2}$. The right-angle vertex will be the center $O$ of $k$, and if we extend the segment $O O_{1}$ until it intersects the circle $k_{1}$ again, we get the point $E_{1}$, which is a point on the circle $k$.

Now suppose that $r_{1}<r_{2}$. Then $O O_{2}<O O_{1}$. Measure the distance $O O_{2}$ from $O$ on the ray $O O_{1}$, and denote the endpoint by $O_{2}^{\prime}$.

In the triangle $O_{1} O_{2} O_{2}^{\prime}$, $O_{1} O_{2}=r_{1}+r_{2}, O_{1} O_{2}^{\prime}=\left(R-r_{1}\right)-\left(R-r_{2}\right)=r_{2}-r_{1}$, and $\angle O_{2} O_{2}^{\prime} O_{1}=135^{\circ}$. Knowing these, the triangle $O_{1} O_{2}^{\prime} O_{2}$ can be constructed: Draw the $135^{\circ}$ arc of the circle with diameter $O_{1} O_{2}$, and cut off the distance $r_{2}-r_{1}$ from $O_{1}$. Thus, we get $O_{2}^{\prime}$. The point $O$ is the vertex of the isosceles right triangle erected on $O_{2} O_{2}^{\prime}$ as the hypotenuse. From here, the circle $k$ can be obtained similarly to the previous cases.

The construction can always be carried out, that is, the tangent circle always exists. As a solution, we always get two circles, which are symmetric with respect to the line $O_{1} O_{2}$.