Adva van két (kívülről vagy belülről) érintkező kör. Szerkesszük meg a közös érintőn azt a pontot, amelyből a két körhöz húzott, egy-egy - a közös érintőtól különböző - érintő egymásra merőleges.

Given are two circles that are tangent (either externally or internally). Construct the point on the common tangent from which the tangents drawn to each circle, distinct from the common tangent, are perpendicular to each other.
Let's first deal with the case when the two circles touch externally. Imagine the problem as solved (see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_6799464417094ef8774dg-1.jpg?height=803&width=775&top_left_y=268&top_left_x=664)

Let $P_{1}$ be a point on one of the common tangents from which a perpendicular tangent can be drawn to each circle. The angle between the perpendicular tangents can be obtained by subtracting the angle between the tangents drawn from $P_{1}$ to the second circle from the angle between the tangents drawn from $P_{1}$ to the first circle. The line $P_{1} O_{1}$ bisects the angle between the tangents drawn from $P_{1}$ to the circle centered at $O_{1}$, and similarly, $P_{1} O_{2}$ bisects the angle between the tangents drawn from $P_{1}$ to the circle centered at $O_{2}$. Therefore, on our diagram, $2 O_{1} P_{1} A_{1} - 2 O_{2} P_{1} A_{1} = 90^{\circ}$ (where $A_{1}$ is the point of tangency of the common tangent with the circle centered at $O_{1}$).

The angle subtended by the segment $O_{1} O_{2}$ at $P_{1}$ can be obtained by subtracting the angle between the bisector of the tangents to the circle centered at $O_{2}$ and the common tangent from the angle between the bisector of the tangents to the circle centered at $O_{1}$ and the common tangent. Thus, the angle subtended by the segment $O_{1} O_{2}$ at $P_{1}$ is $O_{1} P_{1} A_{1} \varangle - O_{2} P_{1} A_{1} \varangle$. This angle is half the angle formed by the tangents drawn from $P_{1}$ to the two circles, which is $45^{\circ}$.

The point $P$ can be constructed by drawing a $45^{\circ}$-angle subtended circle over the segment $O_{1} O_{2}$, and the intersection of this circle with the common tangent is $P_{1}$. The correctness of this construction can be easily verified by reversing the above calculations.

The $45^{\circ}$-angle subtended circle intersects the common external tangent at another point $P_{2}$, the internal tangent at $P_{2}$, and the $135^{\circ}$-angle complementary arc at another point $P_{4}$. Similarly, it can be shown that the tangents drawn from $P_{2}$, $P_{3}$, and $P_{4}$ also form twice the angle that the segment $O_{1} O_{2}$ subtends at these points, so these pairs of tangents are also perpendicular. - If the diagram is reflected across the line $O_{1} O_{2}$, we obtain symmetric points on the other external and internal tangents.

The $45^{\circ}$-angle subtended circle over $O_{1} O_{2}$ always intersects the external tangent. Let the circle centered at $O_{1}$ be the one with the larger radius if the radii are different, so $O_{1} A_{1} \geq O_{2} A_{2}$ (where $A_{1}$ and $A_{2}$ are the points of tangency of the common external tangent). Then $C O_{1} A_{1} \varangle \leq 90^{\circ}$, and the other two angles of the isosceles triangle $C O_{1} A_{1}$ are at least $90^{\circ}$, so $O_{1} A_{1} O_{2} \varangle > O_{1} A_{1} C \varangle \geq 45^{\circ}$. - Therefore, the problem always has 8 solutions.

- If the two circles touch internally, it can similarly be shown that the segment $O_{1} O_{2}$ subtends a $45^{\circ}$ angle at a point $P$ that satisfies the problem. Thus, the construction is the same as before. Here, there is only one external tangent, so if the subtended circle intersects the tangent, there are 4 solutions; if it is tangent, there are 2 solutions; if they have no common points, there are no solutions to the problem.

Note: The method of reasoning used here can also be applied to solve the problem if the tangents form an arbitrary angle $\alpha$ instead of $90^{\circ}$. The $\frac{\alpha}{2}$-angle subtended circle over the segment $O_{1} O_{2}$ intersects the common tangents at points that satisfy the problem.