Question

If the circles \( x^2 + y^2 + 2ax + 2y - 8 = 0 \) and \( x^2 + y^2 - 2x + ay - 14 = 0 \) intersect orthogonally, then the distance between their centers is:

• A. \( \sqrt{242} \)
• B. \( \sqrt{970} \)
• C. \( \sqrt{629} \)
• D. \( \sqrt{541} \)

Hint:

For orthogonal circles, use the intersection condition and solve for unknowns before finding the distance between centers.

Answer 1

The Correct Option is C

Step 1: Find centers and radii Circle 1: \( x^2 + y^2 + 2ax + 2y
- 8 = 0 \) Center \( C_1 = (
-a,
-1) \), radius \( r_1^2 = a^2 + 1^2 + 8 \). Circle 2: \( x^2 + y^2
- 2x + ay
- 14 = 0 \) Center \( C_2 = (1,
-\frac{a}{2}) \), radius \( r_2^2 = 1^2 + \left(\frac{a}{2}\right)^2 + 14 \). Step 2: Orthogonality condition Two circles intersect orthogonally if: \[ 2g_1g_2 + 2f_1f_2 = c_1 + c_2. \] Substituting values from equations, solving for \( a \), and finding the center distance: \[ C_1C_2 = \sqrt{(a + 1)^2 + \left(
-1 + \frac{a}{2}\right)^2}. \] After solving, we get: \[ C_1C_2 = \sqrt{629}. \]