Question

How do you calculate the overlapping area between intersecting circles?

Answer 1

We denote the centres of the smaller circle as $A$ and the larger circle as $B$. The point of their intersection as $C,D$. We take a point P on the intersecting arc of the smaller circle and a point Q on the intersecting arc of the larger circle. The overlapping area is the sum of the areas of the segment CPD and CQD. We subtract the area of triangle ACD from the area of sector ACQD to get the area of segment CPD. Similarly we subtract the area of triangle BCD from area of sector BCPD to get the area of segment CQD in term of radii ${{r}_{1}},{{r}_{2}}$ and common chord $l$.

Complete step by step solution:
Let us draw the diagram as described in the hint with intersecting circles with centres $A,B$ and points of intersection$C,D$. We take a point P on the intersecting arc of the smaller circle with centre $A$ and a point Q on the intersecting arc of the larger circle . $We see that the overlapping area is the sum of the area of the circular segments CPD and CQD.$
Let us find the area of the segment CQD. We need the length of the radii of smaller circle $AC=AD={{r}_{1}}$ .We also need the length of intersecting common chord $CD=l$ and the central angle of the $\angle CAD=\alpha$. $$$$
We know that the area of the sector with central angle $\theta$ is $\dfrac{\theta }{360}\times \pi {{r}^{2}}$. So the area of sector ACQD with central angle $\angle CAD=\alpha$ is
$\text{Area of sector ACQD}=\dfrac{\alpha }{360}\times \pi {{r}_{1}}^{2}$
We denote the point of intersection of line joining centres and the chord as $R$.We know that line segment joining centres perpendicularly bisects the common chord and each other. So we have $CR=DR=\dfrac{l}{2}$. We use Pythagoras theorem in right angled triangle ARC and have

& AR=\sqrt{A{{C}^{2}}-C{{R}^{2}}} \\\ & \Rightarrow AR=\sqrt{{{\left( {{r}_{1}} \right)}^{2}}-{{\left( \dfrac{l}{2} \right)}^{2}}} \\\ \end{aligned}$$ So we have $$\text{Area of }\Delta \text{ACD}=\dfrac{1}{2}\times CD\times AR=\dfrac{1}{2}l\times \sqrt{{{\left( {{r}_{1}} \right)}^{2}}-{{\left( \dfrac{l}{2} \right)}^{2}}}$$ $$\begin{aligned} & \text{Area of segment CQD}=\text{Area of sector ACQD}-\text{Area of }\Delta \text{ACD} \\\ & \Rightarrow \text{Area of segment CQD}=\dfrac{\alpha }{360}\times \pi {{r}_{1}}^{2}-\dfrac{1}{2}l\times \sqrt{{{\left( {{r}_{1}} \right)}^{2}}-{{\left( \dfrac{l}{2} \right)}^{2}}} \\\ \end{aligned}$$ We can find $\alpha $ by taking the tangent of $CAR$ as $$\begin{aligned} & \tan \left( \angle CAR \right)=\dfrac{CR}{AC} \\\ & \Rightarrow \tan \left( \dfrac{\alpha }{2} \right)=\dfrac{\dfrac{l}{2}}{{{r}_{1}}} \\\ & \Rightarrow \dfrac{\alpha }{2}={{\tan }^{-1}}\left( \dfrac{l}{2{{r}_{1}}} \right) \\\ & \Rightarrow \alpha =2{{\tan }^{-1}}\left( \dfrac{l}{2{{r}_{1}}} \right) \\\ \end{aligned}$$ So now the area of circular segment CQD is $$\begin{aligned} & \text{Area of segment CQD}=\dfrac{2{{\tan }^{-1}}\left( \dfrac{l}{2{{r}_{1}}} \right)}{360}\times \pi {{r}_{1}}^{2}-\dfrac{1}{2}l\times \sqrt{{{\left( {{r}_{1}} \right)}^{2}}-{{\left( \dfrac{l}{2} \right)}^{2}}} \\\ & \Rightarrow \text{Area of segment CQD}=\dfrac{2{{\tan }^{-1}}\left( \dfrac{l}{2{{r}_{1}}} \right)}{2\pi }\times \pi {{r}_{1}}^{2}-\dfrac{1}{2}l\times \sqrt{{{\left( {{r}_{1}} \right)}^{2}}-{{\left( \dfrac{l}{2} \right)}^{2}}} \\\ & \Rightarrow \text{Area of segment CQD}={{r}_{1}}^{2}{{\tan }^{-1}}\left( \dfrac{l}{2{{r}_{1}}} \right)-\dfrac{1}{2}l\times \sqrt{{{\left( {{r}_{1}} \right)}^{2}}-{{\left( \dfrac{l}{2} \right)}^{2}}} \\\ \end{aligned}$$ Similarly we can take $\angle CBD=\beta $ and find the area of circular segment CPD as $$\text{Area of segment CPD}={{r}_{2}}^{2}{{\tan }^{-1}}\left( \dfrac{l}{2{{r}_{2}}} \right)-\dfrac{1}{2}l\times \sqrt{{{\left( {{r}_{2}} \right)}^{2}}-{{\left( \dfrac{l}{2} \right)}^{2}}}$$ So the overlapping area is the sum of the areas of circular segment CQD and CPD which is$$\begin{aligned} & A={{r}_{1}}^{2}{{\tan }^{-1}}\left( \dfrac{l}{2{{r}_{1}}} \right)-\dfrac{1}{2}l\times \sqrt{{{\left( {{r}_{1}} \right)}^{2}}-{{\left( \dfrac{l}{2} \right)}^{2}}}+{{r}_{2}}^{2}{{\tan }^{-1}}\left( \dfrac{l}{2{{r}_{2}}} \right)-\dfrac{1}{2}l\times \sqrt{{{\left( {{r}_{2}} \right)}^{2}}-{{\left( \dfrac{l}{2} \right)}^{2}}} \\\ & \Rightarrow A={{r}_{1}}^{2}{{\tan }^{-1}}\left( \dfrac{l}{2{{r}_{1}}} \right)+{{r}_{2}}^{2}{{\tan }^{-1}}\left( \dfrac{l}{2{{r}_{2}}} \right)-\dfrac{1}{2}l\left[ \sqrt{{{\left( {{r}_{1}} \right)}^{2}}-{{\left( \dfrac{l}{2} \right)}^{2}}}+\sqrt{{{\left( {{r}_{2}} \right)}^{2}}-{{\left( \dfrac{l}{2} \right)}^{2}}} \right] \\\ \end{aligned}$$ **Note:** We note that the conclusion $\angle CAR=\angle DAR=\dfrac{\alpha }{2}$ comes from hypotenuse-length congruence of the triangles ACR and ADR. The intersecting areas depend upon the distance between radii $d=\left| {{r}_{2}}-{{r}_{1}} \right|$. If $d=0$ then overlapping area is $\pi r_{1}^{2}$ and if ${{r}_{2}}-{{r}_{1}}>{{r}_{2}}+{{r}_{1}}$ then the area is zero. If the circles are of same radii ${{r}_{2}}={{r}_{1}}=r$ then we have the area as $A=2{{r}^{2}}{{\tan }^{-1}}\left( \dfrac{l}{2r} \right)-l\sqrt{{{r}^{2}}-{{\left( \dfrac{l}{2} \right)}^{2}}}$.