Let the locus of the centre (alpha, beta),bet | vedclass

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(C) Let the centre of the circle be $(\alpha, \beta)$ and its radius be $r$. Since the circle touches the $x$-axis,$r = \beta$.Since it touches the ci

Vedclass · Accepted Answer

$\frac{64}{3}$

Vedclass · Answer

(C) Let the centre of the circle be $(\alpha, \beta)$ and its radius be $r$. Since the circle touches the $x$-axis,$r = \beta$.Since it touches the circle $x^{2} + (y - 1)^{2} = 1$ (centre $(0, 1)$,radius $1$) externally,the distance between the centres equals the sum of the radii:$\sqrt{(\alpha - 0)^{2} + (\beta - 1)^{2}} = \beta + 1$Squaring both sides:$\alpha^{2} + (\beta - 1)^{2} = (\beta + 1)^{2}$$\alpha^{2} + \beta^{2} - 2\beta + 1 = \beta^{2} + 2\beta + 1$$\alpha^{2} = 4\beta$Replacing $(\alpha, \beta)$ with $(x, y)$,the locus $L$ is the parabola $x^{2} = 4y$.The area bounded by $x^{2} = 4y$ and $y = 4$ is given by:$A = 2 \int_{0}^{4} \sqrt{4y} \, dy = 2 \int_{0}^{4} 2 \sqrt{y} \, dy = 4 \left[ \frac{y^{3/2}}{3/2} ight]_{0}^{4} = 4 	imes \frac{2}{3} 	imes 8 = \frac{64}{3}$.

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.

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