Q. 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 :
A) 10
B) 15
C) 12
D) 14
Explanation
[Solution by Dr. Abubacker] $C$ has centre $(-2,2)$ and radius 2; the distance between the centres is $5$. Two intersection points need $|r-2|<5<r+2$, i.e. $3<r<7$. So $3\beta-2\alpha=21-6=15$. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.