Exercise 158

Points $A,B$ are on a circle of radius $13$ and $AB=10$. Point $C$ is the midpoint of the minor arc $AB$. What is $AC$?

Solution

Let $D$ be the midpoint of chord $AB$, so $OD\perp AB$. With $R=13,AB=10$: $AD=5$, and $OD=\sqrt{R^2-AD^2}=\sqrt{169-25}=12$.
Since $C$ is the midpoint of the minor arc, $O,D,C$ are collinear, so $CD=R-OD=13-12=1$.
$AC=\sqrt{AD^2+CD^2}=\sqrt{25+1}=\sqrt{26}$.
The answer is $\boxed{\sqrt{26}}$.