If a straight line y − x = 2 divides the region x 2 + y 2 ≤ 4 into two parts, then the ratio of the area of the smaller part to the area of the greater part is

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Let I be the smaller portion and II be the greater portion of the given figure then,
Area of $I = \int_{- 2}^{0} [4 - x^{2} - (x + 2] d x$ $= [\frac{x}{2} 4 - x^{2} + \frac{4}{2} sin^{- 1} (\frac{x}{2})]_{- 2}^{0} - [\frac{x ^{2}}{2} + 2 x]_{- 2}^{0} = [2 sin^{- 1} (- 1)] - [- \frac{4}{2} + 4] = 2 \times \frac{π}{2} - 2 = π - 2$ Now, area of $II =$ Area of circle $-$ area of $I$ . $= 4 π - (π - 2) = 3 π + 2$ Hence, required ratio $= \frac{area of I}{area of II} = \frac{π - 2}{3 π + 2}$

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