Let the locus of the centre (,), 0, of the circle which touches the circle x^2+(y-1)^2=1 …

Question

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:

Accepted Answer

Radius of circle $S$ touching $x$-axis and centre $(\alpha, \beta)$ is $|\beta|$. According to given conditions

$$
\begin{aligned}
&\alpha^{2}+(\beta-1)^{2}=(|\beta|+1)^{2} \\
&\alpha^{2}+\beta^{2}-2 \beta+1=\beta^{2}+1+2|\beta| \\
&\alpha^{2}=4 \beta \text { as } \beta>0
\end{aligned}
$$

$\therefore \quad$ Required louse is $L: x^{2}=4 y$

The area of shaded region $=2 \int_{0}^{4} 2 \sqrt{y} d y$

$$
\begin{aligned}
&=4 . {\left[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{4} } \\
&=\frac{64}{3} \text { square units. }
\end{aligned}
$$

Solutions · 1