<s>[INST] 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.)
 [/INST] 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:
$$\eta=\frac{\mathcal{A}_1}{\mathcal{A}_2}=\frac{4\text{arccot}\left(\sqrt{7}\right)-\frac{3\sqrt{7}}{8}}{\frac{3 \sqrt{7}}{8}+4 \arctan\left(\sqrt{7}\right)}\approx0.0777567\space\rightarrow\space\eta\approx7.77567\text{%}\tag4$$
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