 Touching Circles [Tangent and Secant Theorem; product of the entire secant and its external part

Two circles with diameters $d$ and $D (d < D)$ are placed inside a semicircle such that each circle touches the arc and the diameter of the semicircle, as well as the other circle. A line is drawn through the centers of the circles, intersecting the extension of the semicircle's diameter at point $M$. A tangent is drawn from point $M$ to the arc of the semicircle ($N$ is the point of tangency). Find $MN$.
Prove that the line passing through the points of tangency of circles with a given semicircle passes through point $M$. Then apply the tangent-secant theorem.

## Solution

Let $O_{1}$ and $O_{2}$ be the centers of the given circles, $r$ and $R$ be their radii respectively $(r<R)$, $A$ and $B$ be the points of tangency with the semicircle, $C$ and $K$ be the points of intersection of the line $AB$ with the first and second circles respectively, $P$ and $Q$ be the points of tangency with the diameter of the semicircle, and $H$ be the center of the semicircle.

Since triangles $AHB$, $K O_{2} B$, and $A O_{1} C$ are isosceles, $\angle ACO_{1} = \angle KBO_{2}$. Therefore, $O_{1}C \parallel O_{2}B$.

Let $L$ be the point of intersection of the lines $AB$ and $O_{1}O_{2}$. Then

$$
\frac{LO_{1}}{LO_{2}} = \frac{CO_{1}}{BO_{2}} = \frac{r}{R}
$$

Therefore,

$$
\frac{LO_{1}}{O_{1}O_{2}} = \frac{r}{R-r}
$$

On the other hand, since $O_{1}P \parallel O_{2}Q$, we have

$$
\frac{MO_{1}}{MO_{2}} = \frac{O_{1}P}{O_{2}Q} = \frac{r}{R}
$$

Therefore,

$$
\frac{MO_{1}}{O_{1}O_{2}} = \frac{r}{R-r} = \frac{LO_{1}}{O_{1}O_{2}}
$$

Thus, points $L$ and $M$ coincide.

By the tangent-secant theorem,

$$
MN^{2} = MB \cdot MA = \frac{R}{r} \cdot MC \cdot MA = \frac{R}{r} \cdot MP^{2}
$$

Let $F$ be the projection of $O_{1}$ onto $O_{2}Q$. From the right triangle $O_{1}FO_{2}$, we find that

$$
O_{1}F = \sqrt{O_{1}O_{2}^{2} - O_{1}F^{2}} = \sqrt{(R+r)^{2} - (R-r)^{2}} = 2\sqrt{rR}
$$

From the similarity of triangles $MPO_{1}$ and $O_{1}FO_{2}$, we have $\frac{MP}{O_{1}F} = \frac{O_{1}P}{O_{2}F}$. Therefore,

$$
MP = \frac{O_{1}F \cdot O_{1}P}{O_{2}F} = \frac{2r\sqrt{rR}}{R-r}
$$

Thus,

$$
MN^{2} = \frac{R}{r} \cdot \left(\frac{2r\sqrt{rR}}{R-r}\right)^{2} = \frac{4R^{2}r^{2}}{(R-r)^{2}}, \quad MN = \frac{2rR}{R-r} = \frac{dD}{D-d}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_7041536e9c9bacd5befeg-21.jpg?height=352&width=1052&top_left_y=1939&top_left_x=503)

Answer

$\frac{Dd}{D-d}$.