A circle with centre \(O\) and radius \(4\) has points \(A\) and \(B\) on its circumference such that \(\angle AOB=90^\circ\).
Another circle with diameter \(AB\) is drawn such that \(O\) lies on its circumference.
Find the area of the shaded region, which is the area inside one circle or the other circle, but not both.
Let \(A_1\) be the region inside the larger circle but outside the smaller circle.
Let \(A_2\) be the region inside the smaller circle but outside the larger circle.
Let \(A_3\) be the region inside sector \(AOB\) but outside of \(\triangle AOB\).
We need to calculate \(A_1 + A_2\).
First, we will calculate \(A_3\).
Since \(\angle AOB = 90^{\circ}\), the
area of sector \(AOB\) is \(\frac{90}{360} = \frac{1}{4}\) the area of
the larger circle.
That is, the area of sector \(AOB\) is
\(\frac{1}{4}\times \pi (4)^2 =
4\pi\).
The area of \(\triangle AOB\) is \(\frac{1}{2}(OA)(OB) = \frac{1}{2}(4)(4) =
8\).
Therefore, \(A_3 =\text{area of sector
}AOB-\text{area of } \triangle AOB = (4\pi - 8)\).
Next we will calculate \(A_2\).
Since \(\angle AOB = 90^{\circ}\), the
Pythagorean theorem tells us \(AB^2 = OA^2 +
OB^2 = 4^2 + 4^2 = 32\). Therefore, \(AB = \sqrt{32} = 4\sqrt{2}\), since \(AB>0\).
Since \(AB\) is a diameter of the
smaller circle, the radius is \(\frac{1}{2}AB
= \frac{1}{2}\left(4\sqrt{2}\right) = 2\sqrt{2}\).
Therefore, \(A_2 + A_3 =
\frac{1}{2}\)(the area of the circle with radius \(2\sqrt{2}\)) \(=
\frac{1}{2}\pi(2\sqrt{2})^2 = \frac{1}{2}\pi(8) = 4\pi\).
Therefore, \(A_2 = 4\pi - A_3 = 4\pi - (4\pi -
8) = 8\).
Finally, we will calculate \(A_1\).
\(A_1\) represents the area inside the
larger circle which is not in the smaller circle.
The larger circle has radius \(4\) and
area \(\pi(4)^2 = 16\pi\).
The smaller circle has radius \(2\sqrt{2}\) and area \(\pi(2\sqrt{2})^2 = 8\pi\). \[\begin{aligned}
A_1 &= \text{(area of larger circle)}- \frac{1}{2}\text{(area of
smaller circle)} - A_3\\
&= 16\pi - \frac{1}{2}(8\pi) - (4\pi - 8)\\
&= 16\pi - 4\pi - 4\pi + 8\\
&= 8\pi + 8
\end{aligned}\] Therefore, the area of the shaded region is equal
to \(A_1 + A_2 = (8\pi + 8) + 8 = (8\pi +
16)\,\text{units}^2.\)