KCET · Maths · Application of Derivatives
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
- C \( 128 \) sq. units
- D \( 8 \) sq. units
Answer & Solution
Correct Answer
(C) \( 128 \) sq. units
Step-by-step Solution
Detailed explanation
\((x+1)^{2}+(y-3)^{2}=64\)
\(x^{2}+x^{2}=(16)^{2}\)
\(2 x^{2}=16.16\)
\(x^{2}=8.16\)
\(x=8 \sqrt{2}\)
Given rectangle is a square
Area of square is \(=8 \sqrt{2} \star 8 \sqrt{2}\)
\(=128 s q\). units
\(x^{2}+x^{2}=(16)^{2}\)
\(2 x^{2}=16.16\)
\(x^{2}=8.16\)
\(x=8 \sqrt{2}\)
Given rectangle is a square
Area of square is \(=8 \sqrt{2} \star 8 \sqrt{2}\)
\(=128 s q\). units
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