Question

If two circles \(x^2 + y^2 - 4x - 2y - 4 = 0\) \& \((x+1)^2 + (y+4)^2 = r^2\) intersect at two distinct points and range of r \(\in (\alpha, \beta)\), then the value of \(\alpha\beta\) is :

Hint:

Remember the geometric conditions for the intersection of two circles with radii \(R_1, R_2\) and distance \(d\) between centers:
Intersect at 2 points: \(|R_1 - R_2|<d<R_1 + R_2\)
Touch externally: \(d = R_1 + R_2\)
Touch internally: \(d = |R_1 - R_2|\)

Answer 1

Correct Answer: 25

Step 1: Find the Centers and Radii of the Circles:
For the first circle, \(C_1: x^2 + y^2 - 4x - 2y - 4 = 0\).
The center is \((-\frac{-4}{2}, -\frac{-2}{2}) = (2, 1)\).
The radius is \(R_1 = \sqrt{(-2)^2 + (-1)^2 - (-4)} = \sqrt{4+1+4} = \sqrt{9} = 3\).
For the second circle, \(C_2: (x+1)^2 + (y+4)^2 = r^2\).
The center is \((-1, -4)\).
The radius is \(R_2 = r\).

Step 2: Find the Distance Between the Centers:
The distance \(d\) between the centers \(C_1(2, 1)\) and \(C_2(-1, -4)\) is: \[ d = \sqrt{(-1-2)^2 + (-4-1)^2} = \sqrt{(-3)^2 + (-5)^2} = \sqrt{9+25} = \sqrt{34} \]
Step 3: Apply the Condition for Intersection:
Two circles intersect at two distinct points if the distance between their centers is greater than the absolute difference of their radii and less than the sum of their radii. \[ |R_1 - R_2|<d<R_1 + R_2 \] Substituting the values we found: \[ |3 - r|<\sqrt{34}<3 + r \]
Step 4: Solve the Inequality for r:
We have two conditions to solve from the inequality: 1. \(\sqrt{34}<3 + r \Rightarrow r>\sqrt{34} - 3\)
2. \(|3 - r|<\sqrt{34}\)
This implies \(-\sqrt{34}<3 - r<\sqrt{34}\).
From \(3 - r<\sqrt{34}\), we get \(r>3 - \sqrt{34}\). (Since r must be positive, this is always true as \(3-\sqrt{34}\) is negative).
From \(-\sqrt{34}<3 - r\), we get \(r-3<\sqrt{34}\), which means \(r<3 + \sqrt{34}\). Combining both conditions, we get: \[ \sqrt{34} - 3<r<\sqrt{34} + 3 \] This is the range of r, so it corresponds to the interval \((\alpha, \beta)\).
Step 5: Calculate the Final Value:
We have \(\alpha = \sqrt{34} - 3\) and \(\beta = \sqrt{34} + 3\). We need to find the value of \(\alpha\beta\). \[ \alpha\beta = (\sqrt{34} - 3)(\sqrt{34} + 3) \] Using the difference of squares formula \((a-b)(a+b) = a^2 - b^2\): \[ \alpha\beta = (\sqrt{34})^2 - 3^2 = 34 - 9 = 25 \]