Use calculus to find the area bounded by the circle $x^2+y^2-2x-2y-23=0$ and the pair of lines $x^2+2xy+y^2-7x-7y+12=0$

Question

Use calculus to find the area bounded by the circle $x^2+y^2-2x-2y-23=0$ and the pair of lines $x^2+2xy+y^2-7x-7y+12=0$.

I tried to solve the two equations by subtracting them.

$2xy+35=5x+5y$,on further solving it is getting too complicated and not giving solutions.Hence could not move ahead.Please help me.

Answer 1

The lines are $x+y=3$ and $x+y=4$. These are parallel lines. The equation of the circle is $(x-1)^2+(y-1)^2=25$. Hence you may rotate the lines so that they are parallel to the $x$-axis. You may also place the origin at the center of the circle. The distance between these lines is $1/\sqrt2$ and the distance between $x+y=3$ from the center of the circle is $1/\sqrt2$. Hence the area is $$S=2\int_{1/\sqrt2}^{2/\sqrt2}\sqrt{25-y^2}dy=-\frac{7}{2}+\sqrt{46}-25 \arcsin\left(\frac{1}{5 \sqrt{2}}\right)+25 \arcsin \left(\frac{\sqrt{2}}{5}\right)$$

Before rotation:

After rotation and centering:

Answer 2

HINT

Do you have to use calculus?

The circle is $(x-1)^2+(y-1)^2=5^2$ and the pair of straight lines are $(x+y-3)(x+y-4)=0$, so you could find where each line intersects the circle separately. A picture would help, and some trigonometry.