A circle with centre \(O\) has diameter \(AB\). A line segment is drawn from a point \(C\) on the circumference of the circle to \(D\) on \(OB\) such that \(CD\perp OB\) and \(CD=2\) units. Two circles are drawn on \(AB\). One has diameter \(AD\) and the other has diameter \(DB\).
Determine the area of the shaded region. That is, determine the area inside the circle centred at \(O\) but outside of the circle with diameter \(AD\) and outside of the circle with diameter \(DB\).
Note: In solving this problem, it may be helpful to use the fact that the angle inscribed in a circle by the diameter is \(90\degree\). For example, in the following diagram, \(PQ\) is a diameter and \(\angle PRQ\) is inscribed in the circle by diameter \(PQ\). Therefore, \(\angle PRQ=90\degree.\)
Let the radius of the circle with diameter \(DB\) be \(r\). Then \(DB =
2r\). Let the radius of the circle with diameter \(AD\) be \(R\). Then \(AD=2R\).
Also, \(AB = AD + DB = 2R + 2r\), and
so the radius of the circle with centre \(O\) is \(R+r.\)
It follows that the area of the circle with diameter \(AD\) is \(\pi R^2\), the area of the circle with diameter \(DB\) is \(\pi r^2\), and the area of the circle with centre \(O\) is \(\pi (R+r)^2.\)
To determine the shaded area, we calculate the area of the circle with centre \(O\) and subtract the area of the circle with diameter \(AD\) and the area of the circle with diameter \(DB\). That is, \[\begin{aligned} \text{Shaded Area}&=\pi (R+r)^2-\pi R^2 -\pi r^2\\ &=\pi (R^2+2Rr+r^2)-\pi R^2-\pi r^2\\ &=\pi R^2 +2\pi Rr+\pi r^2-\pi R^2-\pi r^2\\ &=2\pi Rr \end{aligned}\]
Join \(A\) to \(C\) and \(C\) to \(B\). Since \(AB\) is a diameter and \(\angle ACB\) is inscribed in a circle by that diameter, we know that \(\angle ACB = 90\degree\).
Since \(CD\perp OB\), then \(\angle ODC = \angle ADC = \angle BDC = 90\degree\). We will use the Pythagorean Theorem in the three triangles \(\triangle ADC\), \(\triangle BDC\), and \(\triangle ACB\), to establish a relationship between \(R\) and \(r.\)
In \(\triangle ADC\), \(AC^2=AD^2+CD^2=(2R)^2+2^2=4R^2+4\).
In \(\triangle BDC\), \(BC^2=DB^2+CD^2=(2r)^2+2^2=4r^2+4\).
In \(\triangle ACB\), \(AB^2=AC^2+BC^2=(4R^2+4)+(4r^2+4)=4R^2+4r^2+8\).
But \(AB^2=(AD+DB)^2=(2R+2r)^2=4R^2+8Rr+4r^2.\)
Therefore, \(4R^2+8Rr+4r^2=4R^2+4r^2+8\) and \(8Rr=8\) or \(Rr=1\) follows.
Thus, the shaded area is equal to \(2\pi Rr = 2\pi (1) =2\pi\text{ units}^2.\)
Note: The relationship \(Rr=1\) could also be established using similar triangles as follows:
In \(\triangle ADC\), \(\angle CAD+\angle ACD=90\degree\). Also, since \(\angle ACB=90\degree\), \(\angle ACD+\angle DCB=90\degree\).
Thus, \(\angle CAD+\angle ACD = \angle ACD+\angle DCB\), which simplifies to \(\angle CAD=\angle DCB.\)
Now \(\angle CAD=\angle DCB\) and \(\angle CDA = \angle CDB=90\degree\).
Therefore, \(\triangle ADC \sim \triangle CDB\) by Angle Angle Angle (AAA) similarity.
From triangle similarity, \(\dfrac{AD}{CD}=\dfrac{CD}{DB}\), and so \(\dfrac{2R}{2}=\dfrac{2}{2r}\) and \(Rr=1\) follows.