The maximum area of a rectangle inscribed in | vedclass

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(C) The given equation of the circle is $(x+1)^{2}+(y-3)^{2}=64$. Comparing this with the standard form $(x-h)^{2}+(y-k)^{2}=r^{2}$,we get the radius

Vedclass · Accepted Answer

$128 	ext{ sq. units}$

Vedclass · Answer

(C) The given equation of the circle is $(x+1)^{2}+(y-3)^{2}=64$. Comparing this with the standard form $(x-h)^{2}+(y-k)^{2}=r^{2}$,we get the radius $r = \sqrt{64} = 8$. For a rectangle inscribed in a circle,the maximum area is achieved when the rectangle is a square. The diagonal of this square is equal to the diameter of the circle,which is $d = 2r = 2 	imes 8 = 16$. Let the side of the square be $a$. Then,by Pythagoras theorem,$a^{2} + a^{2} = d^{2}$. $2a^{2} = 16^{2} = 256$. $a^{2} = 128$. The area of the square is $a^{2} = 128 	ext{ sq. units}$.

The maximum area of a rectangle inscribed in the circle $(x+1)^{2}+(y-3)^{2}=64$ is

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