Q58 / 120 · VITEEE 2010 · Mathematics

The line joining $(5,0)$ to $(10 \cos \theta, 10 \sin \theta)$ is divided internally in the ratio $2: 3$ at $\mathrm{P}$. If $\theta$ varies, then the locus of $\mathrm{P}$ is

  • A
    a straight line
  • B
    a pair of straight lines
  • Correct answer
    C
    a circle
  • D
    None of the above

Solution

Let coordinates of $P$ be $(h, k)$, then $$ \begin{array}{l} h=\frac{2(10 \cos \theta)+3(5)}{2+3}=4 \cos \theta+3 \\ \text { and } k=\frac{2(10 \sin \theta)+3(0)}{2+3}=4 \sin \theta \end{array} $$ [Using the internal section formula] $$ \Rightarrow \frac{h-3}{4}=\cos \theta \text { and } \frac{k}{4}=\sin \theta $$ Squaring and adding both of these equations, $$ \begin{array}{l} \frac{(h-3)^{2}}{16}+\frac{k^{2}}{16}=\cos ^{2} \theta+\sin ^{2} \theta \\ \Rightarrow(h-3)^{2}+k^{2}=16 \end{array} $$ Therefore, locus of point $P$ is $(x-3)^{2}+y^{2}=16$ which is a circle.

Hence, the correct answer is option: 'C'.

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