Question

A circle has equation (x-a)^2 +(y-a)^2=a^2 where a is a constant.t. The line y+x-a=0 splits the area of the circle into 2 parts, A1
and A2 where A1>A2. Find the area of A2
giving your answer in the form ((a^2)/b)*(c*pi + d) where b c, and d are integers.

Answer 1

Answer on Question #53405 - Math - Analytic Geometry

Question

A circle has equation $(x-a)^2 + (y-a)^2 = a^2$ where $a$ is a constant. The line $y+x-a=0$ splits the area of the circle into 2 parts, A1 and A2 where A1>A2. Find the area of A2 giving your answer in the form $((a^2)/b)^*(c*pi + d)$ where $b, c,$ and $d$ are integers.

Solution

A2 is red and the right triangle is shown by means of black thick segments in figure. Area of A2 equals circular sector area minus area of the right triangle:

$$S(A2) = \frac{\pi a^2}{4} - \frac{1}{2} a^2 = \frac{a^2}{2} \left(\frac{\pi}{2} - 1\right)$$

Answer: $\frac{a^2}{2} \left(\frac{\pi}{2} - 1\right)$

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