# Euclidean Geometry - Circles Problem Solving

This wiki is about problem solving on circles. You need to be familiar with some (if not all) theorems on circles.

#### Contents

## Sample Problems

Find the area of the shaded region in the above diagram.

The area of the shaded region is equal to the area of the semicircle with diameter $3$ plus the area of the semicircle with diameter $4$ plus the area of the right triangle minus the area of the semicircle with diameter $5$. We have

$\begin{aligned} \text{Area}&=\dfrac{\pi}{4}\big(3^2\big)+\dfrac{\pi}{4}\big(4^2\big)+\dfrac{1}{2}(3)(4)-\dfrac{\pi}{4}(5)^2\\\\ &=\dfrac{9}{4}\pi + 4\pi + 6 - \dfrac{25}{4}\pi\\\\ &=6.\ _\square \end{aligned}$

$BC$ is the diameter of a circle with center at $O$. If $\angle ABO=30^\circ$ and $\angle OAD=40^\circ$, find the measure of $\angle COD$.

$AC$. Then by Thales' theorem, $\triangle BAC$ is a right triangle.

Draw lineSince $\triangle BOA$ is isosceles with $OB=OA=\text{radius}$, $\angle OAB=\angle ABO=30^\circ.$

So, $\angle CAD=90^\circ-30^\circ-40^\circ=20^\circ.$

By the inscribed angle theorem,

$\angle COD=2(\angle CAD)=2(20^\circ)=40^\circ.\ _\square$

$DC$ is the diameter of the circle with center at $A.$ $BE$ is perpendicular to $DC$. If $DE=15$ and $EC=9$, find $BE.$

$\triangle DEB \sim \triangle DBC$, we have

Since$\dfrac{BE}{DE}=\dfrac{BC}{BD}\implies \dfrac{BE}{15}=\dfrac{BC}{BD}\implies BE=15\dfrac{BC}{BD}\qquad (1)$

and

$\dfrac{BE}{BD}=\dfrac{BC}{DC}\implies \dfrac{BE}{BD}=\dfrac{BC}{24}\implies BE=\dfrac{(BD)(BC)}{24}. \qquad (2)$

Substituting (1) into (2) gives

$\begin{aligned} 15\dfrac{BC}{BD}&=\dfrac{(BD)(BC)}{24}\\ 15(24)&=(BD)^2\\ BD&=\sqrt{360}\\&=6\sqrt{10}. \end{aligned}$

Using the Pythagorean theorem, we have

$BC=\sqrt{DC^2-BD^2}=\sqrt{24^2-\big(6\sqrt{10}\big)^2}=\sqrt{216}=6\sqrt{6}.$

Finally,

$BE=15\dfrac{BC}{BD}=15\left(\dfrac{6\sqrt{6}}{6\sqrt{10}}\right)=\dfrac{15\sqrt{6}}{\sqrt{10}}.$

Rationalizing the denominator gives

$BE=\dfrac{15\sqrt{6}}{\sqrt{10}} \cdot \dfrac{\sqrt{10}}{\sqrt{10}}=\dfrac{30}{10}\sqrt{15}=3\sqrt{15}.\ _\square$

## Challenge Problems

$BC$ is tangent to both a circle with center at $A$ and a circle with center at $D$. The area of the circle with center at $A$ is $225\pi$ and the area of the circle with center at $D$ is $36\pi$.

If $BC=16$, find the distance between the centers of the two circles.

## See Also

**Cite as:**Euclidean Geometry - Circles Problem Solving.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/euclidean-geometry-circles-problem-solving/