Q116 / 180 · KCET 2018 · Mathematics

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

A
$ 64 \mathrm{sq} . $ units
B
$ 72 \mathrm{sq} . $ units
Correct answer
C
$ 128 $ sq. units
D
$ 8 $ sq. units

Solution

$(x+1)^{2}+(y-3)^{2}=64$ $x^{2}+x^{2}=(16)^{2}$ $2 x^{2}=16.16$ $x^{2}=8.16$ $x=8 \sqrt{2}$ Given rectangle is a square Area of square is $=8 \sqrt{2} \star 8 \sqrt{2}$ $=128 s q$. units

Hence, the correct answer is option: 'C'.

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

Solution

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