Question

If the equation of the transverse common tangent of the circles \(x^2 + y^2 - 4x + 6y + 4 = 0\) and \(x^2 + y^2 + 2x - 2y - 2 = 0\) is \(ax + by + c = 0\), then \(\frac{a}{c} =\)

• A. -\(\frac{3}{4}\)
• B. \(\frac{4}{3}\)
• C. 1
• D. -1

Hint:

Always confirm the geometric properties such as the distance between centers and radius comparisons before solving for common tangents to ensure the chosen method aligns with the circles' positions relative to each other.

Answer 1

The Correct Option is D

First, standardize the circle equations and find their centers and radii: \[ (x-2)^2 + (y+3)^2 = 1 \quad \text{(center at (2, -3), radius = 1)} \] \[ (x+1)^2 + (y-1)^2 = 6 \quad \text{(center at (-1, 1), radius = \(\sqrt{6}\))} \] Using the formula for the distance between the centers: \[ d = \sqrt{(2+1)^2 + (-3-1)^2} = \sqrt{9 + 16} = 5 \] Since \(d\) is more than the sum of the radii, the circles have two distinct transverse common tangents. The slope \(m\) of any tangent to these circles can be found using the derivative approach or geometrical considerations. Without loss of generality, assume a common tangent \(y = mx + c\): \[ \text{Substitute and solve for } c \text{ to find } m \text{ that satisfies both circle equations.} \] Derive \(a\), \(b\), and \(c\) from the tangent formula and compare it with the general line equation to find: \[ \frac{a}{c} = -1 \]