JEE Mains · Maths · STD 12 - 8. Application and integration
Let the locus of the centre \((\alpha, \beta), \beta>0\), of the circle which touches the circle \(x ^{2}+( y -1)^{2}=1\) externally and also touches the \(x\)-axis be \(L\). Then the area bounded by \(L\) and the line \(y =4\) is.
- A \(\frac{32 \sqrt{2}}{3}\)
- B \(\frac{40 \sqrt{2}}{3}\)
- C \(\frac{64}{3}\)
- D \(\frac{32}{3}\)
Answer & Solution
Correct Answer
(C) \(\frac{64}{3}\)
Step-by-step Solution
Detailed explanation
\((\alpha-0)^{2}+(\beta-1)^{2}=(\beta+1)^{2}\) \(\alpha^{2}=4\,\beta\) \(x ^{2}=4\,y\) \(A =2 \int_{0}^{4}\left(4-\frac{ x ^{2}}{4}\right) dx =\frac{64}{3}\)
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