An isosceles right triangle has vertices (0,0), (4,0) and (0,4). A circle of radius 2 is centred at (2,2). Find the area common to both.

Step-by-step solution

1. 1

   Write both regions as inequalities. The triangle is $x\ge0$, $y\ge0$, $x+y\le4$; the disc is

   $(x-2)^2+(y-2)^2\le 4$

   The overlap is the set satisfying all four conditions at once, so each of the three triangle constraints has to be examined separately against the disc.

2. 2

   Test the hypotenuse constraint. The centre $(2,2)$ satisfies $2+2=4$, so the centre lies exactly on the line $x+y=4$. A line through the centre of a circle is a diameter line, cutting the disc into two congruent halves.

   $\text{area of disc with }x+y\le4=\tfrac12\pi(2)^2=2\pi$

3. 3

   Test the two leg constraints. The distance from $(2,2)$ to the $x$-axis is $2$, and to the $y$-axis is $2$ - both equal to the radius. So the circle is tangent to each axis (touching at $(2,0)$ and $(0,2)$) and never crosses either one.

   The disc therefore lies entirely in $x\ge0,\ y\ge0$, and the legs remove nothing.

4. 4

   Combine the three tests. Only the hypotenuse cuts the disc, and it halves it. The intersection is exactly the half-disc on the side $x+y\le4$:

   $\text{Area}=\tfrac12\cdot\pi r^2=\tfrac12\cdot 4\pi=2\pi$

5. 5

   Sanity-check the magnitude. The triangle has area $\tfrac12(4)(4)=8$ and the full disc has area $4\pi\approx12.57$. The overlap $2\pi\approx6.283$ is less than both, and less than half the triangle - consistent with a half-disc tucked against the hypotenuse ✓.