JEE MAIN - Mathematics (2022 - 25th July Morning Shift - No. 13)
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 $$\mathrm{L}$$. Then the area bounded by $$\mathrm{L}$$ and the line $$y=4$$ is:
$$
\frac{32 \sqrt{2}}{3}
$$
$$
\frac{40 \sqrt{2}}{3}
$$
$$\frac{64}{3}$$
$$
\frac{32}{3}
$$
Explanation
_25th_July_Morning_Shift_en_13_1.png)
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 $L: x^{2}=4 y$
_25th_July_Morning_Shift_en_13_2.png)
The area of shaded region $=2 \int_{0}^{4} 2 \sqrt{y} d y$
$$ \begin{aligned} &=4 . {\left[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{4} } \\\\ &=\frac{64}{3} \text { square units. } \end{aligned} $$
Comments (0)
