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

Question

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.

Answer 1

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.

Answer 2

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)..