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.

If I may refer to sketch of Omran Kouba, you  need even not rotate each pink segment by $\pm \tan ^{-1} \frac12$, you can straightaway use the standard formula of area of circular segment ( subtract sector and triangle areas)..
