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…

Question

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 :

Aicharya · Accepted Answer

$\begin{aligned} & C_1 C_2=\sqrt{(2+2)^2+(5-2)^2} \ & =\sqrt{16+9} \ & =5 \ & r+2>5 \ & r>3 \ & r < 5+2 \ & r < 7 \ & herefore \alpha=3, \beta=7 \ & 3 \beta-2 \alpha=3(7)-2(3)=21-6=15\end{aligned}$

A circle $C$ of radius 2 lies in the second quadrant and touches both the coordinate axes. Let $r$ be the…

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