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 a center 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 $(α, β)$ , then 3 $β$ – 2 $α$ is equal to

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Detailed Solution

$\begin{aligned} S_{1} : (x + 2)^{2} + (y - 2)^{2} = 2^{2} \\ S_{2} : (x - 2)^{2} + (y - 5)^{2} = r^{2} \end{aligned}$

Both circles intersect at two points

$\begin{aligned} ∴ |r_{1} - r_{2}| < c_{1} c_{2} < r_{1} + r_{2} \end{aligned}$

$\begin{array}{r} | r - 2 | < 5 < 2 + r \end{array}$

$\begin{array}{r} \Rightarrow 3 < r < 7 \end{array}$

$\begin{array}{r} r \in (3, 7) \end{array}$

$\begin{array}{r} α = 3, β = 7 \end{array}$

$3 β - 2 α = 15$

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