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  • Two circles are inscribed and circumscribed about a square A B C D, length of each side of the square is 16. P and Q are two points respectively on these circles, then <img alt="\Sigma(P

Two circles are inscribed and circumscribed about a square A B C D, length of each side of the square is 16. P and Q are two points respectively on these circles, then

\Sigma(P A)^2+\Sigma(Q A)^2 is equal to

Option: 1

1720        


Option: 2

 1712 


Option: 3

1792


Option: 4

1710


Answers (1)

best_answer

Let the centre of the square be the origin O and the lines through O parallel to the sides of the square be the coordinate axis. Then the vertices of the square are

A(8,8), B(-8,8), C(-8,-8) \text { and } D(8,-8) \text {. }
The radii of the inscribed and circumscribed circles are respectively 8 and O A=\sqrt{\left(8^2+8^2\right)}=8 \sqrt{2} and their centre is at the origin.
Let the coordinate of P\: be\: the\: (8 \cos \theta, 8 \sin \theta)and that of Q \: be\: (8 \sqrt{2} \cos \phi, 8 \sqrt{2} \sin \phi)
Then\: \Sigma(P A)^2=8^2\left[(\cos \theta-1)^2+(\sin \theta-1)^2+(\cos \theta+1)^2\right.+(\sin \theta-1)^2
\begin{aligned} +(\cos \theta+1)^2 & \left.+(\sin \theta+1)^2-(\cos \theta-1)^2+(\sin \theta+1)^2\right] & =2 \times 8^2\left[2 \cos ^2 \theta+2+2 \sin ^2 \theta+2\right] \end{aligned}

=12 \times 8^2\\Similarly\: \Sigma(Q A)^2=16 \times 8^2\\ \therefore \Sigma(P A)^2+\Sigma(Q A)^2=28 \times 8^2=1792

 

 

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