C1C2=(2+2)2+(5−2)2=16+9=5r+2>5r>3r<5+2r<7∴α=3,β=73β−2α=3(7)−2(3)=21−6=15
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 (α,β), then 3β−2α is equal to :
Held on 22 Jan 2025 · Verified 6 Jul 2026.
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