Problem of the Week Problem E and Solution Another Circle

Problem

Points $$A$$ and $$B$$ are on a circle with centre $$O$$ and radius $$6$$ cm, such that $$\angle AOB=60 \degree$$.

Determine the radius of the circle which passes through points $$A,~B,$$ and $$O$$.

Solution

Let $$C$$ be the centre of the circle that passes through $$A$$, $$B$$, and $$O$$. Then $$CA$$, $$CB$$, and $$CO$$ are radii. Therefore, $$CA=CB=CO=r$$, where $$r$$ is the radius of the circle.

In $$\triangle CAO$$ and $$\triangle CBO$$, $$CA=CB$$, $$CO$$ is common, and $$OA=OB$$. Therefore, $$\triangle CAO \cong \triangle CBO$$ and it follows that $$\angle COA=\angle COB$$. But $$\angle AOB=60\degree$$. Therefore, $$\angle COA=\angle COB=30\degree$$.

In $$\triangle CAO$$, $$CA=CO=r$$ and $$\triangle CAO$$ is isosceles. Therefore, $$\angle CAO=\angle COA=30\degree$$ and $$\angle ACO=180\degree-30\degree-30\degree=120\degree$$.

From here we can find the length of $$r$$ using either the sine law or the cosine law.

Method 1: Using the sine law, \begin{aligned} \dfrac{CA}{\sin{(\angle COA)}}&=\dfrac{OA}{\sin{(\angle ACO)}}\\ \dfrac{r}{\sin{30\degree}}&=\dfrac{6}{\sin{120\degree}}\\ r&=\dfrac{6}{\sin{120\degree}}\times \sin{30\degree}\\ r&=\dfrac{6}{\frac{\sqrt{3}}{2}}\times \frac{1}{2}\\ r&=6\times \frac{2}{\sqrt{3}} \times \frac{1}{2}\\ r&=\frac{6}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\\ r&=2\sqrt{3}\text{ cm}\end{aligned} Therefore, the radius of the circle that passes through $$A,~B,$$ and $$O$$ is $$2\sqrt{3}$$ cm.

Method 2: Using the cosine law, \begin{aligned} CA^2&=CO^2+AO^2-2\times CO\times AO\times \cos{(\angle COA)}\\ r^2&=r^2+6^2-2(6)(r)\cos{30\degree}\\ 12r\cos{30\degree}&=36\\ r\cos{30\degree}&=3\\ r\times \frac{\sqrt{3}}{2}&=3\\ r\times \sqrt{3}&=6\\ r\times \sqrt{3}\times \sqrt{3}&=6\times \sqrt{3}\\ 3r&=6 \sqrt{3}\\ r&=2 \sqrt{3}\text{ cm}\end{aligned} Therefore, the radius of the circle that passes through $$A,~B,$$ and $$O$$ is $$2\sqrt{3}$$ cm.