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  • Let the locus of the cetre , of the circle which touches the circle

Let the locus of the cetre \mathrm{(\alpha, \beta), \beta>0}, of the circle which touches the circle \mathrm{x^{2}+(y-1)^{2}=1} externally and also touches the x-axis be \mathrm{L}. Then the area bounded by \mathrm{L} and the line \mathrm{y= 4} is :

Option: 1

\frac{32 \sqrt{2}}{3}


Option: 2

\frac{40 \sqrt{2}}{3}


Option: 3

\frac{64}{3}


Option: 4

\frac{32}{3}


Answers (1)

best_answer

\mathrm{(\alpha-0)^{2}+(\beta-1)^{2}=(\beta+1)^{2}} \\

\mathrm{ \alpha^{2}=4 \beta} \\

\mathrm{x^{2}=4 y} \\

\mathrm{A= 2 \int_{0}^{4}\left(4-\frac{x^{2}}{4}\right) d x} \\

\mathrm{= 2\left[4 x-\frac{x^{3}}{12}\right]_{0}^{4}} \\

=\frac{64}{3}

Hence correct option is 3

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