Three of the four vertices of a rectangle are (5,11), (16,11) and (16,-2). What is the area of the intersection of this rectangular region and the region inside the graph of the equation (x - 5)^2 + (y + 2)^2 = 16? Express your answer as a common fraction in terms of pi.

Guest Oct 28, 2020

#1**0 **

https://www.desmos.com/calculator/p8bel3908b = Link to diagram

In a circle, "the center-radius form of the circle equation is in the format $(x – h)^2 + (y – k)^2 = r^2,$ with the center being at the point (h, k) and the radius being r." So, $(x - 5)^2 + (y + 2)^2 = 16$ is an equation of a circle centered at $(5,2).$ However, our rectangle has a **vertex at the center!** Thus, the area of the intersection will be $\frac{1}{4} \cdot 16 \cdot \pi,$ or $4\pi.$

Pangolin14 Oct 28, 2020