+0

# coordinate geometry

0
118
1

What is the area of the portion of the circle defined by x^2 - 12x + y^2 = 28 that lies above the x-axis and to the right of the line y = 8 - x?

Jan 28, 2021

#1
+8461
+1

We first find the intersection point of x^2 - 12x + y^2 = 28 and y = 8 - x.

$$\begin{cases} x^2 - 12x + y^2 = 28\\ y = 8-x \end{cases}$$

After some work, we get (x, y) = $$(7-\sqrt{31}, 1+\sqrt{31})$$. We don't need the other intersection point.

We also get that the region is represented by the compound inequality $$\max(0, 8- x)\leq y \leq \sqrt{28 + 12x - x^2}$$.

We use the straight line x = 8 to divide the region into 2 parts.

Using integration, the area is

$$\displaystyle \int^{8}_{7-\sqrt{31}}\left(\sqrt{28 +12x - x^2}-(8-x)\right)\,dx+\int^{14}_8 \sqrt{28 + 12x - x^2}\,dx$$

which would be $$16\pi + 32\arcsin\dfrac{\sqrt{31} - 1}{8} - \sqrt{31} - 1 \approx 63.14\text{ sq.units}$$

The computation is too long so I didn't include it.

Here's a verification of my answer by Wolfram Alpha:

WolframAlpha computation

This is the diagram I used:

Desmos diagram

Jan 28, 2021