A circle C of radius 2 lies in the second quadrant and touches both coordinate axes.

Question

A circle $C$ of radius $2$ lies in the second quadrant and touches both 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:

Accepted Answer

$S_1:(x+2)^2+(y-2)^2=2^2$

$$ \mathrm{S}_2:(\mathrm{x}-2)^2+(\mathrm{y}-5)^2=\mathrm{r}^2 $$

Both circle intersect at two points

$$ \begin{aligned} & \therefore\left|\mathrm{r}_1-\mathrm{r}_2\right|<\mathrm{c}_{\mathrm{c}} \mathrm{c}_2<\mathrm{r}_1+\mathrm{r}_2 \ & |\mathrm{r}-2|<5<2+\mathrm{r} \ & \Rightarrow 3<\mathrm{r}<7 \ & \mathrm{r} \in(3,7) \ & \alpha=3, \beta=7 \ & 3 \beta-2 \alpha=15 \end{aligned} $$

Solutions · 1