Paragraph:
Read the following passage and answer the questions.
Let \(A B C D\) be a square of side length 2 units. \(C_2\) is the circle through vertices \(A, B, C, D\) and \(C_1\) is the circle touching all the sides of square \(A B C D\). \(L\) is the line through \(A\).Question:
If \(P\) is a point on \(C_1\) 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

\[
\text { Here, equation of } C_2:(x-1)^2+(y-1)^2=(\sqrt{2})^2 \text { and } C_1:(x-1)^2+(y-1)^2=(1)^2
\]

\[
\begin{aligned}
& \therefore P(1+\cos \theta, 1+\sin \theta) \\
& \text { and } Q(1+\sqrt{2} \sin \theta, 1+\sqrt{2} \sin \theta) \\
& \therefore P A^2+P B^2+P C^2+P D^2 \\
& =\left\{(1+\cos \theta)^2+(1+\sin \theta)^2\right\} \\
& +\left\{(\cos \theta-1)^2+(1+\sin \theta)^2\right\} \\
& +\left\{(1+\cos \theta)^2+(1-\sin \theta)^2\right\} \\
& +\left\{(1-\cos \theta)^2+(1-\sin \theta)^2\right\} \\
& =16 \\
& \therefore \quad \frac{\Sigma Q A^2}{\Sigma P A^2}=\frac{12}{16}=0.75 \\
&
\end{aligned}
\]
Hence (a) is the correct answer.