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

To find the interval of values for \( r \) such that the circles \( (x + 1)^2 + (y + 4)^2 = r^2 \) and \( x^2 + y^2 - 4x - 2y - 4 = 0 \) intersect at two distinct points, we need to determine the conditions for intersection.

Step 1: Identify the centers and radii of the circles.

• Circle 1: \( (x + 1)^2 + (y + 4)^2 = r^2 \).
  • Center: \((-1, -4)\)
  • Radius: \(r\)
• Circle 2: \( x^2 + y^2 - 4x - 2y - 4 = 0 \).
  • Rewriting the equation in standard form: \(\begin{align*} x^2 - 4x + y^2 - 2y &= 4 \\ (x-2)^2 - 4 + (y-1)^2 - 1 &= 4 \\ (x-2)^2 + (y-1)^2 &= 9 \end{align*}\)
  • Center: \((2, 1)\)
  • Radius: \(3\)

Step 2: Calculate the distance between the centers.

The distance \( d \) between the centers \((-1, -4)\) and \((2, 1)\) is given by:

\(d = \sqrt{(2 - (-1))^2 + (1 - (-4))^2} = \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}\)

Step 3: Determine the conditions for intersection at two distinct points.

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

\(|r - 3| < \sqrt{34} < r + 3\)

Thus, we solve the inequalities:

1. \( r + 3 > \sqrt{34} \)
2. \( r - 3 < \sqrt{34} \)

This gives:

• From \( r + 3 > \sqrt{34} \):
• From \( r - 3 < \sqrt{34} \):

Hence, \( r \) is in the interval:

\(\left(\sqrt{34} - 3, \sqrt{34} + 3\right)\)

Step 4: Find \( \alpha\beta \).

Given the interval \( (\alpha, \beta) = \left(\sqrt{34} - 3, \sqrt{34} + 3\right) \), calculate:

\(\alpha = \sqrt{34} - 3\)

\(\beta = \sqrt{34} + 3\)

The product \(\alpha\beta\) is:

\((\sqrt{34} - 3)(\sqrt{34} + 3) = (\sqrt{34})^2 - 3^2 = 34 - 9 = 25\)