Through point $N$, two lines tangent to a certain circle with center $O$ are drawn. On one of these lines, point $A$ is taken, and on the other line, point $B$ is taken such that $O A=O B, O A>O N$. It is known that $N A=a$, $N B=b, O A=c(a \neq b)$. Find $O N$.
Let the circle touch the lines $N A$ and $N B$ at points $P$ and $Q$ respectively. Then the circle is inscribed in the angle $P N Q$, and points $A$ and $B$ lie on the sides of one of the two angles adjacent to the angle $P N Q$. Prove that the segments $A P$ and $B Q$ are equal and express them in terms of $a, b$, and $c$.

## Solution

Let the circle touch the lines $N A$ and $N B$ at points $P$ and $Q$ respectively. Then the circle is inscribed in the angle $P N Q$. Note that since $O A=O B$ and

$O A>O N$, the points $A$ and $B$ lie on the sides of one of the two angles adjacent to the angle $P N Q$. The right triangles $A O P$ and $B O Q$ are equal by the leg and hypotenuse, so $A P=B Q$. From the equality $N P=N Q$ it follows that $N A-A P=B Q-N B$, or $a-A P=B Q-b$, and since $A P=B Q$, then

$A P=B Q=1 / 2(a+b)$. Therefore, $N P=A N-A P=a-1 / 2(a+b)=1 / 2(a-b)$. From the right triangles $A O P$ and NOP we sequentially find that

$O P^{2}=O A^{2}-A P^{2}=c^{2}-1 / 4(a+b)^{2}, O N^{2}=N P^{2}+O P^{2}=1 / 4(a-b)^{2}+c^{2}-1 / 4(a+b)^{2}=c^{2}-a b$.

![](https://cdn.mathpix.com/cropped/2024_05_06_39480ed8b0d60d88e837g-26.jpg?height=503&width=586&top_left_y=946&top_left_x=745)

## Answer

$\sqrt{c^{2}-a b}$