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The line joining (5, 0) to $(10 \cos θ, 10 \sin θ)$ is divided internally in the ratio 2:3 at P, then the locus of P is

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$(x - 3)^{2} + y^{2} = 16$

$x^{2} = y - 3$

$x^{2} + 2 x y + y^{2} - 6 x = 0$

$x + y - 3 = 0$

answer is C.

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Detailed Solution

$\begin{array}{l} P (x, y) = \frac{2 (10 \cos θ, 10 \sin θ) + 3 (5, 0)}{2 + 3} \\ \Rightarrow x = \frac{20 \cos θ + 15}{5} \Rightarrow y = \frac{20 \sin θ}{5} \end{array}$

$\Rightarrow$ $(x - 3)^{2} + y^{2} = 16$

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