Circle 2 Question 17
18. If $P$ is a point of $C _1$ and $Q$ is a point on $C _2$, then $\frac{P A^{2}+P B^{2}+P C^{2}+P D^{2}}{Q A^{2}+Q B^{2}+Q C^{2}+Q D^{2}}$ is equal to
(a) 0.75
(b) 1.25
(c) 1
(d) 0.5
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Solution:
- Let the, equation of circles are
$$ C _1:(x-1)^{2}+(y-1)^{2}=(1)^{2} $$
and $C _2:(x-1)^{2}+(y-1)^{2}=(\sqrt{2})^{2}$
$\therefore$ Coordinates of $P(1+\cos \theta, 1+\sin \theta)$
$$ \begin{aligned} & \text { and } \quad Q(1+\sqrt{2} \cos \theta, 1+\sqrt{2} \sin \theta) \\ & \therefore \quad P A^{2}+P B^{2}+P C^{2}+P D^{2} \\ & ={(1+\cos \theta)^{2}+(1+\sin \theta)^{2} }+{(\cos \theta-1)^{2}+(1+\sin \theta)^{2} } \\ & +{(\cos \theta-1)^{2}+(\sin \theta-1)^{2} } \\ & +{(1+\cos \theta)^{2}+(\sin \theta-1)^{2} }=12 \end{aligned} $$
Similarly, $Q A^{2}+Q B^{2}+Q C^{2}+Q D^{2}=16$
$$ \therefore \quad \frac{\Sigma P A^{2}}{\Sigma Q A^{2}}=\frac{12}{16}=0.75 $$
