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Two circles are inscribed and circumscribed about a square ABCD, length of each side of the square is 32 . P and Q are two points respectively on these circles, then \mathrm{ \Sigma(P A)^2+\Sigma(Q A)^2} is equal to

Option: 1

7168


Option: 2

7000


Option: 3

7050


Option: 4

7150


Answers (1)

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Let the centre of the square be the origin O and the lines through O parallel to the side of the square be the coordinate axes. Then the vertices of the square are A(16, 16), B(–16, 16), C(–16, –16) and D(16, –16). The radii of the inscribed and circumscribed circles are
respectively 16 and \mathrm{O A=\sqrt{16^2+16^2}=16 \sqrt{2}} and their centre is at the origin.

\mathrm{\text { Let the coordinate of } P \text { be }(16 \cos \theta, 16 \sin \theta) \text { and that of }}

\mathrm{Q \text { be }(16 \sqrt{2} \cos \phi, 16 \sqrt{2} \sin \phi)}

\mathrm{\begin{aligned} & \text { then } \Sigma(P A)^2=16^2\left[(\cos \theta-1)^2+(\sin \theta-1)^2+(\cos \theta+1)^2\right. \\ & +(\sin \theta-1)^2+(\cos \theta+1)^2+(\sin \theta+1)^2 \\ & \left.+\quad+(\cos \theta-1)^2+(\sin \theta+1)^2\right] \\ & =2 \times 16^2\left[2 \cos ^2 \theta+2+2 \sin ^2 \theta+2\right]=12 \times(16)^2 \end{aligned}}

\mathrm{\begin{aligned} & \text { similarly } \Sigma(Q A)^2=16 \times(16)^2 \\ & \therefore \quad \Sigma(P A)^2+\Sigma(Q A)^2=28 \times(16)^2=7168 \end{aligned}}

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Pankaj Sanodiya

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