A) a straight line
B) a pair of straight lines
C) a circle
D) None of the above
Correct Answer: C
Solution :
Let coordinate of P are (h, k) then \[h=\frac{2(10cos\theta )+3(5)}{2+3}=4\cos \theta +3\] and \[k=\frac{2\,(10\sin \theta )+3\,(0)}{2+3}=4\sin \theta \] \[\left[ \begin{align} & \text{Using}\,\text{the}\,\text{internal}\,\text{section}\,\text{formula,} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\begin{matrix} h=\frac{{{m}_{1}}{{x}_{2}}+{{m}_{2}}{{x}_{1}}}{{{m}_{1}}+{{m}_{2}}} \\ k=\frac{{{m}_{1}}{{y}_{2}}+{{m}_{2}}{{y}_{1}}}{{{m}_{1}}+{{m}_{2}}} \\ \end{matrix} \\ \end{align} \right]\] \[\Rightarrow \] \[\frac{h-3}{4}=\cos \theta \]and \[\frac{k}{4}=\sin \theta \] Squaring and adding both of these equations \[\frac{{{(h-3)}^{2}}}{16}+\frac{{{k}^{2}}}{16}={{\cos }^{2}}\theta +{{\sin }^{2}}\theta \] \[\Rightarrow \] \[{{(h-3)}^{3}}+{{k}^{2}}=16\] Therefore, locus of point P is \[{{\left( x-3 \right)}^{2}}+{{y}^{2}}=16\] which is a circle.
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