Question

Let the set of all values of \( r \), for which the circles \( (x + 1)^2 + (y + 4)^2 = r^2 \) and \( x^2 + y^2 - 4x - 2y - 4 = 0 \) intersect at two distinct points be the interval \( (\alpha, \beta) \). Then \( \alpha\beta \) is equal to

• A. 25
• B. 21
• C. 24
• D. 20

Hint:

Two circles intersect at two distinct points if the distance between their centers lies strictly between the sum and the absolute difference of their radii.

Answer 1

The Correct Option is A

The given circles are \[ (x + 1)^2 + (y + 4)^2 = r^2 \] and \[ x^2 + y^2 - 4x - 2y - 4 = 0. \] Step 1: Find the center and radius of each circle. 
For the first circle, \[ (x + 1)^2 + (y + 4)^2 = r^2, \] the center is \[ C_1(-1, -4) \] and the radius is \[ r_1 = r. \] For the second circle, rewrite the equation by completing squares: \[ x^2 - 4x + y^2 - 2y = 4, \] \[ (x - 2)^2 + (y - 1)^2 = 9. \] Thus, the center is \[ C_2(2, 1) \] and the radius is \[ r_2 = 3. \] Step 2: Find the distance between the centers. 
The distance between \( C_1(-1, -4) \) and \( C_2(2, 1) \) is \[ d = \sqrt{(2 + 1)^2 + (1 + 4)^2} = \sqrt{3^2 + 5^2} = \sqrt{34}. \] Step 3: Condition for intersection at two distinct points. 
Two circles intersect at two distinct points if \[ |r_1 - r_2|<d<r_1 + r_2. \] Substituting the values, \[ |r - 3|<\sqrt{34}<r + 3. \] Step 4: Solve the inequalities. 
From \[ \sqrt{34}<r + 3, \] we get \[ r>\sqrt{34} - 3. \] From \[ |r - 3|<\sqrt{34}, \] we get \[ -\sqrt{34}<r - 3<\sqrt{34}, \] which gives \[ r<3 + \sqrt{34}. \] Hence, the interval is \[ (\alpha, \beta) = (\sqrt{34} - 3,\; \sqrt{34} + 3). \] Step 5: Find \( \alpha\beta \). 
\[ \alpha\beta = (\sqrt{34} - 3)(\sqrt{34} + 3) = 34 - 9 = 25. \]