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 :
(A)15✓ Correct
(B)14
(C)12
(D)10
Answer: (A)
Solution
A circle of radius 2 lying in the second quadrant and touching both axes must have its center at (−2,2). Call this C:
S1:(x+2)2+(y−2)2=4.
Let the second circle have center (2,5) and radius r:
S2:(x−2)2+(y−5)2=r2.
Distance between centers:
d=(2−(−2))2+(5−2)2=16+9=5.
Two circles intersect at exactly two points precisely when
∣r1−r2∣<d<r1+r2,i.e.∣r−2∣<5<r+2.
From 5<r+2: r>3. From ∣r−2∣<5: −3<r<7.
Combined with r>3, the valid range is r∈(3,7), so α=3,β=7.
3β−2α=21−6=15.
JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Single correct option
