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Q)

A rectangular hyperbola whose cetre is C is cut by any circle of radius r in four points $P,Q,R$ and S. Then $CP^2+CQ^2+CR2+CS^2$ is equal to.

$\begin{array}{1 1}(A)\;r^2 \\(B)\;2r^2 \\(C)\;3\;r^2 \\(D)\;4\;r^2 \end{array}$

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A)
let the equation of the rectangular hyperbola be $xy=c^2$-----(1)
and equation of circle be $x^2+y^2=r^2$------(2)
Put $ y= \large\frac{c^2}{x}$ in (2) we get,
$x^2 +\large\frac{c^4}{c^2} $$=r^2$
$x^4-r^2x^2+c^4=0$------(3)
Now, $CP^2+CQ^2+CR^2+CS^2$
=> $x_1^2+y_1^2+x_2^2+y_2^2+x_3^2+y_3^2+x_4^2+y_4^2$
=> $ \bigg( \sum \limits_{i=1}^{4} x_i \bigg)^2 - 2\sum x_1x_2 + $ $\bigg( \sum \limits_{i=1}^{4} y_i \bigg)^2 -2 \sum y_1y_2$
=> $ 2r^2+2r^2 $ from (3)
=> $4r^2$
Hence D is the correct answer.
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