Area between circle equation and a line

Question

Find the area of common region expressed by two inequalities $x^2+y^2 \leq 4$ and $x+y \leq -2$.

It is needed in phi constant, so I assume I need to change it into its polar equation, then substitute the line equation into the circle one. But I don't know how to get the meeting point of the inequalities.

Answer 1

I think your first inequality should be $x^2 + y^2 \leq 4$. What you have right now for that inequality is basically all of $\mathbb{R}^2$ outside a circle of radius 2, which doesn't make sense. You can rearrange $x + y \leq -2$ into an expression for $x$ or $y$, then substitute it into the circle equation to find the points of intersection.

Answer 2

The line intersects the circle at (-2,0) and (0, -2), because they satisfy

$$x^2+y^2=4$$ $$x+y=-2$$

It is easier then to visualize that the common region is just the difference between the quarter circle of area $\pi$ and the overlapping right triangle of area 2.

Thus, the area of the shaded region is $\pi - 2$.