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Prove It It always helps to draw a picture.
You can have the second circle be centred anywhere on the first, I've just chosen the point $\displaystyle (x, y) = (1, 0)$.
It's important to note that the equations of the circles are
$\displaystyle x^2 + y^2 = 1$ and $\displaystyle (x - 1)^2 + y^2 = \frac{9}{4}$.
By subtracting the first equation from the second we find
$\displaystyle (x - 1)^2 + y^2 - (x^2 + y^2) = \frac{9}{4} - 1$
$\displaystyle x^2 - 2x + 1 + y^2 - x^2 - y^2 = \frac{5}{4}$
$\displaystyle -2x + 1 = \frac{5}{4}$
$\displaystyle -2x = \frac{1}{4}$
$\displaystyle x = -\frac{1}{8}$.
So the equations intersect at $\displaystyle x = -\frac{1}{8}$.
Putting the equations as $\displaystyle y$ in terms of $\displaystyle x$ gives:
$\displaystyle y = \pm \sqrt{1 - x^2}$ and $\displaystyle y = \pm \sqrt{\frac{9}{4} - (x - 1)^2}$.
Since these are symmetrical about the $\displaystyle x$ axis, we can work out the areas enclosed by the above semicircles, and then double it.
So we can finally say:
$\displaystyle A = 2\left(\int_{-\frac{1}{2}}^{-\frac{1}{8}}{\sqrt{\frac{9}{4} - (x - 1)^2}\,dx} + \int_{-\frac{1}{2}}^{1}{\sqrt{1 - x^2}\,dx}\right)$
You will need to use trigonometric substitution to solve these.
Try $\displaystyle x - 1 = \frac{3}{2}\sin{\theta}$ for the first
and $\displaystyle x = \sin{\theta}$ for the second.