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The larger area bounded by $y^{2}$ = 4x and $x^{2} + y^{2} - 2 x - 3 = 0$ is

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$2 π - \frac{4}{3}$

$2 π + \frac{4}{3}$

$2 π + \frac{8}{3}$

$2 π + \frac{2}{3}$

answer is C.

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Detailed Solution

$x^{2} + y^{2} - 2 x - 3 = 0$

$\Rightarrow (x - 1)^{2} + y^{2} = 4$

It meets y 2 = 4x at x = 1.

$∴$ Required area $= 2 \int_{0}^{1} \sqrt{4 x} d x$ + area of the semicircle

$= \frac{8}{3} + 2 π$

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