The area of the region described by a moving point P lying inside the circle ${(x - a)}^{2} + y^{2} = a^{2}$ above x–axis such that its radial distance, (i.e., distance of the point P from circle along the radius through P), is greater than its distance from x–axis is $K a^{2}$ , then the value of 6K is

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answer is 4.

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Detailed Solution

Let us take point P(h, k) on the boundary of the region described by the point under consideration, then
$P B = k \Rightarrow P A = a - k$
$\Rightarrow {(h - a)}^{2} + k^{2} = {(a - k)}^{2}$
$\Rightarrow k = h - \frac{h^{2}}{2 a}$
$\Rightarrow y = x - \frac{x^{2}}{2 a}$ is boundary of the region
So required area will be
$A = \int_{0}^{2 a} (x - \frac{x^{2}}{2 a}) d x$
= ${(\frac{x^{2}}{2} - \frac{x^{3}}{6 a})}_{0}^{2 a} = \frac{4 a^{2}}{2} - \frac{8 a^{3}}{6 a} = 2 a^{2} - \frac{4 a^{2}}{3} = \frac{2 a^{2}}{3}$