Find the percentage of the area of each circle that overlaps. 
The diagram shows two overlapping circles with centres $X$ and $Y$. Both circles have radius $r$ cm and the distance between the centres, $XY$, is $1.5r$ cm. 
I got the answer $18.2$% (to $3$s.f), but the answer key states otherwise ($14.4$%).
My Steps are:


*

*Find the area of the circle ($\pi r^2$)

*Find the area of $AXBY$ ($r^2$)

*Subtract sector $XAB$ to get the area of $ABY$ ($ABY=(r^2) - (\pi r^2)/4$)

*Area of $AXBY$ - $2$(Area of $ABY$)= Area of overlap = $(r^2)((\pi/2)-1)$

*Percentage= $((\pi-2)/2\pi)x100$% = $18.2$% ($3$s.f.)
 A: AXBY isn't a square! Thus the area isn't $r^2$.
For this method to work, you also want to make sure that line AY doesn't "clip" the circle on the right (indeed it doesn't, but it would've if the two circles were within $\sqrt{2}r$.
A: Let's assume that $\text{r}=1$ and we will work through the solution and then the OP can generalize it.

Well, we know that the equation of a circle is given by:
$$\left(x-\text{a}\right)^2+\left(\text{y}-\text{b}\right)^2=\text{r}^2\tag1$$
Where $\left(\text{a},\text{b}\right)$ are the center coordinates of the circle and $\text{r}$ is the radius of the circle.
So, in your case, we have two circles so we write $1=\text{r}_1=\text{r}_2$ and we have $\left(\text{a}_1,\text{b}_1\right)$ and $\left(\text{a}_2,\text{b}_2\right)$. Using Mathematica I used the following code:
ContourPlot[{(x + (-3/4))^2 + (y + 0)^2 == 
   1^2, (x + (3/4))^2 + (y + 0)^2 == 1^2}, {x, -2, 2}, {y, -2, 2}, 
 GridLines -> {{0, 3/4, -3/4}, {0}}]

And it gave me:

So, we have $\left(\text{a}_1,\text{b}_1\right)=\left(-\frac{3}{4},0\right)$ and $\left(\text{a}_2,\text{b}_2\right)=\left(\frac{3}{4},0\right)$.
The surface area of the part where the circle's overlap can be found using:
$$\mathcal{A}_1:=4\int_0^\frac{1}{4}\frac{\sqrt{\left(1-4x\right)\left(7+4x\right)}}{4}\space\text{d}x=4\text{arccot}\left(\sqrt{7}\right)-\frac{3\sqrt{7}}{8}\tag2$$
And the total area of both the circle's is given by:
$$\mathcal{A}_2:=2\int_{-\frac{3}{4}-1}^{\frac{3}{4}+1}\left(\text{K}_1+\text{K}_2\right)\space\text{d}x=\frac{3 \sqrt{7}}{8}+4 \arctan\left(\sqrt{7}\right)\tag3$$
Where $\text{K}_1=\theta\left(x\right)\cdot\frac{\sqrt{\left(7-4x\right)\left(1+4x\right)}}{4}$ and $\text{K}_2=\theta\left(-x\right)\cdot\frac{\sqrt{\left(1-4x\right)\left(7+4x\right)}}{4}$.
So, for the percentage we get:
$$\eta=\frac{\mathcal{A}_1}{\mathcal{A}_2}=\frac{4\text{arccot}\left(\sqrt{7}\right)-\frac{3\sqrt{7}}{8}}{\frac{3 \sqrt{7}}{8}+4 \arctan\left(\sqrt{7}\right)}\approx0.0777567\space\rightarrow\space\eta\approx7.77567\text{%}\tag4$$
