# Finding area enclosed by two curves (circles)

Given information : F : r=18cosθ, rg=6,G : r=6,rf=9

The question is: Find the area of region A in Figure 1. I know that I need to subtract the smaller area enclosed by the two circles. I tried finding the intersection points first to do this, but the angle was complicated, so I am not sure if I need to use the angle to solve this problem...

18cosθ = 6
θ=cos-1(1 ⁄ 3)

Do I use this as the angle? Also, the picture is small and I can't see how the intersection lines would make the smaller area into two areas (dividing the top part from the bottom) or not. I am really confused with how to interpret this problem.

• You can use calculus to first find area of common region and then subtract it from circle area – King Tut Mar 11 '18 at 4:21
• This can be done without calculus using geometry also. Common chord divides common area I to two parts which can be found separately – King Tut Mar 11 '18 at 4:23
• @KingTut I have tried that, but the answer was wrong. When I am trying to find the area of the common region, I assumed that the intersection line aligns with the line of the bigger circle within the area enclosed by the two circles. Do you think this generalization is correct? – h.jb Mar 11 '18 at 17:33
• I think you mean whether circle intersect orthogonally. They do not necessarily, so it's not a good assumption. To find common area, divide it into two parts by common chord. Now try to find area of these two segments separately by subtracting area of triangle from area of sector.. – King Tut Mar 11 '18 at 18:46
• @KingTut I am not sure what you mean by that.. Could you show me how you would calculate that? – h.jb Mar 12 '18 at 1:56

Draw a common chord of circles $G$ and $F$. The plan is to calculate the areas of segments $CEDB$ and $CEDO$ separately.

Let the distance of this chord from centre of $G$ be $x$. Upon using pythagorean theorem on $\triangle CEO$ and $\triangle CEA$, we will get $x = 2$.

To calculate the area of segment $CEDB$, we subtract the area of triangle $COD$ from sector $CODB$. Note that $CE = \sqrt{36-x^2} = 4\sqrt{2}$. So angle $COD$ can be written as $2 \arctan (\tfrac{CE}{OE}) = 2 \arctan{2\sqrt2}$.

Area of sector $CODB$ is simply $\frac{2\arctan(2\sqrt 2)}{2\pi} \pi \cdot 6^2 = 36 \arctan(2\sqrt 2)$. The area of triangle $COD$ is $8\sqrt 2$.

Therefore area of segment $CEDB$ is $\color{red}{36 \arctan(2\sqrt 2)-8\sqrt 2}$

Repeat the calculations to get area of segment $CEDO$. That is subtract area of triangle $CDA$ from area of sector $CODA$. I get upon calculation, area of segment $CEDO$ as $\color{red}{81\arctan(\tfrac{4\sqrt 2}{ 7})-28\sqrt 2}$.

Now area of the shaded region in question is simply the area of circle $F$ minus the area of segments $CEDB$ and $CEDO$. So the required answer is:

$$81 \pi-36\arctan(2\sqrt2) - 81 \arctan(\tfrac{4\sqrt2}{7}) + 36\sqrt 2$$