If a straight line y−x=2 divides the region x2+y2≤4 into two parts, then the ratio of the area of the smaller part to the area of the greater part is:
Solutions · 1
Official solution
The chord y=x+2 cuts the circle at (−2,0) and (0,2).
\[\text{Area}_{\text{smaller}} = \int_{-2}^0\left[\sqrt{4-x^2}-(x+2)\right]dx = \left[\frac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\frac x2\right]_{-2}^0 - \left[\frac{x^2}{2}+2x\right]_{-2}^0\]
\[= \left[0-(-\pi)\right] - \left[0-(-2)\right] = \pi-2\]
Total circle area =4π, so the greater part =4π−(π−2)=3π+2.
\[\text{Ratio} = (\pi-2):(3\pi+2)\]