Find the area of the circle that falls between the circle $x^2+y^2=5$ and the lines $x^2-4y^2+6x+9=0$ 
Find the area of the circle that falls between the circle $x^2+y^2=5$ and the lines $x^2-4y^2+6x+9=0$.

I tried to solve this question. The lines are $x-2y+3=0$ and $x+2y+3=0$ which intersect at $(-3,0)$ and the circle has its center at $(0,0)$ and radius $\sqrt5$. The figure was asymmetric and calculations were difficult. So I shifted the origin to $(-3,0)$. In the new system the equations become $X+2Y=0$ and $X-2Y=0$ and the circle equation becomes $(X-3)^2+Y^2=5$. Now I found the point of intersection of lines and the circle which comes out to be $(\frac{4}{5},\frac{2}{5})$
Now when I integrate and required area is 4 times the integral $\int_{0}^{4/5}\frac{x}{2}dx+\int_{4/5}^{\sqrt5}\sqrt{5-(x-3)^2}dx$
but the answer $5(\pi-\arcsin\frac{4}{5})+\frac{24}{5}$ is still elusive and the calculations have not simplified even after origin shifting. Is the origin shifting futile. What is the correct way to solve it? Please help me.

The line of equation $x-2y+3=0$ intersects the circle of equation $x^2+y^2=5$ at
$A(-\frac{11}{5},-\frac{2}{5})$ and $B(1,2)$. It follows that
$AB=\frac{8}{\sqrt{5}}$.
$\qquad\qquad\qquad$
Using the notation of the figure we have $\sin\theta=\frac{AB}{2OA}=\frac{4}{5}$. That is $\theta=\arcsin\frac{4}{5}$. Moreover, $OH^2=5-\frac{16}{5}=\frac{9}{5}$, or $OH=\frac{3}{\sqrt5}$.
Now, $${\cal S}=\frac{\theta}{2}OA^2-\frac{1}{2}AB\cdot OH
=\frac{5}{2}\arcsin\frac{4}{5}-\frac{4}{\sqrt5}\frac{3}{\sqrt5}=
\frac{5}{2}\arcsin\frac{4}{5}-\frac{12}{5}$$
Finally the desired area is 
$${\frak A}=\pi(\sqrt5)^2-2{\cal S}=5\left(\pi-
\arcsin\frac{4}{5}\right)+\frac{24}{5}.$$
Which is the right answer.
