JEE MAIN - Mathematics (2025 - 22nd January Morning Shift - No. 9)
A circle C of radius 2 lies in the second quadrant and touches both the coordinate axes. Let r be the radius of a circle that has centre at the point $(2,5)$ and intersects the circle $C$ at exactly two points. If the set of all possible values of r is the interval $(\alpha, \beta)$, then $3 \beta-2 \alpha$ is equal to :
10
12
14
15
Explanation
_22nd_January_Morning_Shift_en_9_1.png)
$$ S_1:(x+2)^2+(y-2)^2=2^2 $$
$$ \mathrm{S}_2:(\mathrm{x}-2)^2+(\mathrm{y}-5)^2=\mathrm{r}^2 $$
Both circle intersect at two points
$$ \begin{aligned} & \therefore\left|\mathrm{r}_1-\mathrm{r}_2\right|<\mathrm{c}_{\mathrm{c}} \mathrm{c}_2<\mathrm{r}_1+\mathrm{r}_2 \\ & |\mathrm{r}-2|<5<2+\mathrm{r} \\ & \Rightarrow 3<\mathrm{r}<7 \\ & \mathrm{r} \in(3,7) \\ & \alpha=3, \beta=7 \\ & 3 \beta-2 \alpha=15 \end{aligned} $$
Comments (0)
