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# Three of the four vertices of a rectangle are (5,11), (16,11) and (16,-2).

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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.

Oct 28, 2020

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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 $4\pi.$