Two circles touching at P

The figure shows two circles, centres \(O\) and \(C\), radii \(r_1\) and \(r_2\), which touch externally at \(P\). Given that \(A\hat{O}P=\dfrac{\pi}{4}\) radians, \(B\hat{C}P = \dfrac{\pi}{3}\) radians and \(AB\) is parallel to \(OPC\), show that \(\dfrac{r_1}{r_2} = \dfrac{\sqrt{6}}{2}\).

Two circles touching at P with right-angled triangles OAD and CBE

Dropping perpendiculars from \(A\) and \(B\) gives us the above diagram. Thus we have

\[ r_1 \sin \dfrac{\pi}{4} = AD= BE = r_2 \sin \frac{\pi}{3}. \]

And so we have \[\dfrac{r_1}{r_2} = \dfrac{\sin \dfrac{\pi}{3}}{\sin \dfrac{\pi}{4}} = \dfrac{\left(\dfrac{\sqrt{3}}{2}\right)}{\left(\dfrac{1}{\sqrt{2}}\right)} = \dfrac{\sqrt{6}}{2}.\]

Hence find the ratio of the areas of the sectors \(OAP\), \(CBP\).

The area of a sector containing an angle \(\theta\) and with radius \(r\) is \(\dfrac{1}{2}r^2 \theta\). So \[\begin{align*} A_1 &=\dfrac{1}{2}r_1^2 \times \dfrac{\pi}{4}=\dfrac{\pi r_1^2}{8},\\ A_2 &=\dfrac{1}{2}r_2^2 \times \dfrac{\pi}{3}=\dfrac{\pi r_2^2}{6}.\\ \end{align*}\]

Hence \[\frac{A_1}{A_2} = \dfrac{6}{8} \left( \dfrac{r_1}{r_2} \right)^2 = \dfrac{9}{8}.\]