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:

• A. 15
• B. 14
• C. 12
• D. 10

Hint:

For two circles to intersect at two points, the distance between their centers must be greater than the difference of their radii and less than the sum of their radii.

Answer 1

The Correct Option is A

The equation of circle C with radius 2 and center \( (-2, 2) \) is: \[ S_1: (x + 2)^2 + (y - 2)^2 = 2^2 \] The equation of the circle with center \( (2, 5) \) and radius \( r \) is: \[ S_2: (x - 2)^2 + (y - 5)^2 = r^2 \] For both circles to intersect at exactly two points, the distance between the centers must satisfy: \[ |r_1 - r_2| <c_1c_2 <r_1 + r_2 \] Where \( r_1 = 2 \) and \( r_2 = r \), the distance between centers is \( 5 \): \[ |r - 2| <5 <r + 2 \] \[ 3 <r <7 \] Thus, \( r \in (3, 7) \). Therefore, \( \alpha = 3 \) and \( \beta = 7 \). \[ 3\beta - 2\alpha = 3 \times 7 - 2 \times 3 = 21 - 6 = 15 \]

Answer 2

A circle of radius \(2\) tangent to both axes in the second quadrant has centre \[ A = (-2,\,2),\qquad R=2. \]

The other circle has centre \[ B=(2,\,5), \] and radius \(r\). The distance between centres is \[ d=|AB|=\sqrt{(2-(-2))^2+(5-2)^2}=\sqrt{4^2+3^2}=5. \]

Two circles intersect in exactly two distinct points iff \[ |R-r|

Solve the inequalities: \[ |2-r|<5 \implies -5<2-r<5 \implies -33. \] Combining gives \(3

Compute: \[ 3\beta-2\alpha=3\cdot7-2\cdot3=21-6=15. \]

Answer

15 (Option 1)