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:
- (1) 15
- (2) 14
- (3) 12
- (4) 10
Solution:
Circle C lies in the second quadrant and touches both axes with radius 2.
So, its centre is $C_1(-2, 2)$ and radius $r_1 = 2$.
The second circle has centre $C_2(2, 5)$ and radius $r_2 = r$.
Distance between centres $C_1 C_2$:
$$ d = \sqrt{(2 – (-2))^2 + (5 – 2)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 $$
For two circles to intersect at exactly two points, the condition is:
$$ |r_1 – r_2| < d < r_1 + r_2 $$
$$ |2 – r| < 5 < 2 + r $$
Taking the second part:
$$ 5 < 2 + r \Rightarrow r > 3 $$
Taking the first part:
$$ |2 – r| < 5 \Rightarrow -5 < 2 – r < 5 $$
$$ -7 < -r < 3 \Rightarrow -3 < r < 7 $$ Since radius $r > 0$, combining with $r > 3$, we get:
$$ r \in (3, 7) $$
So, $\alpha = 3$ and $\beta = 7$.
Value required: $3\beta – 2\alpha = 3(7) – 2(3) = 21 – 6 = 15$.
Ans. (1)