Two equal circles, each of radius \(\quantity{10}{cm}\), have centres at \(A\) and \(B\) respectively and the length of \(AB\) is \(\quantity{17.4}{cm}\). The circles intersect at \(C\) and \(D\). Find

  1. the angle \(CBA\) and express it in radians correct to three significant figures,
The prior diagram, with the line segment $CD$ and the angle $\theta = \angle CBA$ added.

The line segment \(CD\) is the perpendicular bisector of \(AB\), and so we have the following right-angled triangle.

The isolated right-angled triangle.

Thus, \[ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\quantity{8.7}{cm}}{\quantity{10}{cm}} = 0.87 \] and so \[ \theta = \arccos(0.87) = \text{$0.51559\!\ldots$ radians} \] which, to three significant figures, is \(0.516\) radians.

  1. the area of the minor sector \(BCD\),

By symmetry, \(\theta = \angle CBA = \angle DBA\), and so \[ \angle DBC = \angle DBA + \angle CBA = 2\theta = 2 \arccos(0.87). \]

The area of a sector of radius \(r\) and enclosed angle \(\theta\) (in radians) is \(\dfrac{1}{2}r^2\theta\), so the area of the sector \(CDB\) is \(\dfrac{1}{2}10^2(2 \arccos(0.87)) = 51.559...,\)

which to \(3\)sf is \(\quantity{51.6}{cm^2}.\)

  1. the area common to both circles.
A diagram that isolates the region of the minor sector $BCD$ that is not part of the triangle $BCD$.

By symmetry, the area common to both circles is twice the area of this shaded region, and the area of this shaded region is equal to \[ \text{area of the minor sector $BCD$} - \text{area of $\triangle BCD$}. \]

The area of a triangle is \(\dfrac{1}{2}ab \sin C\), so the triangle \(BCD\) has area \(\dfrac{1}{2}100 \sin (2\arccos(0.87))=42.895...\)

Thus the area of the shaded region is \(51.559... - 42.895... = 8.664...,\) and so the area shared between the circles is (\(3\)sf) \(\quantity{17.3}{cm^2}.\)