[ Pythagorean Theorem (direct and inverse). ] [ Common tangent to two circles ]

Two circles touch each other externally. Find the length of their common external tangent (between the points of tangency), if the radii are 16 and 25.
Drop a perpendicular from the center of the smaller circle to the radius of the larger circle, extended to the point of tangency with the common external tangent.

## Solution

Let $O$ and $O_{2}$ be the centers of the given circles, $M$ and $M_{2}$ be the points of tangency with the common external tangent, and $K$ be the foot of the perpendicular dropped from point $O$ to $O_{2} M_{2}$. By the Pythagorean theorem, $O K^{2}=O_{2} O^{2}-O_{2} M^{2}=(25+16)^{2}-(25-16)^{2}=4 \cdot 16 \cdot 25=40^{2}$

Therefore, $M M_{2}=O K=40$.

![](https://cdn.mathpix.com/cropped/2024_05_06_14311972f49418f31424g-13.jpg?height=446&width=626&top_left_y=2215&top_left_x=727)

## Answer

40. 

The radii of two circles are 27 and 13, and the distance between their centers is 50. Find the lengths of their common tangents.

## Solution

Let $O$ and $O_{2}$ be the centers of the circles with radii 27 and 13, $A$ and $B$ be the points of tangency of the circles with their common external tangent, $C$ and $D$ be the points of tangency with the internal tangent, and $O_{2} P$ be the perpendicular dropped from $O_{2}$ to $O A$ (left figure). By the Pythagorean theorem, $O_{2} P^{2}=O_{2} O^{2}-O P^{2}=50^{2}-(27-13)^{2}=48^{2}$, so $A B=O_{2} P=48$.

Drop a perpendicular $O Q$ from $O$ to the extension of the radius $O_{2} D$ (right figure). Then $C D^{2}=O Q^{2}=O_{2} O^{2}-O_{2} Q^{2}=50^{2}-(27+13)^{2}=30^{2}$

![](https://cdn.mathpix.com/cropped/2024_05_06_14311972f49418f31424g-14.jpg?height=366&width=1612&top_left_y=660&top_left_x=223)

## Answer

48 and 30.