Question

The product of the slopes of the common tangents drawn to the circles $x^2+y^2+2x-2y-2=0$ and $x^2+y^2-2x+2y+1=0$ is

• A. $-1$
• B. $3$
• C. $-\frac{8}{3}$
• D. $1$

Hint:

Common Tangents.
Use distance = radius condition for each circle. If line $y = mx + c$ touches both, eliminate $c$ and form quadratic in $m$. Product of slopes = product of roots.

Answer 1

The Correct Option is D

Find centers and radii by completing the square: Circle 1: $(x+1)^2 + (y-1)^2 = 4 \Rightarrow \text{center } C_1 = (-1,1),\ r_1 = 2$
Circle 2: $(x-1)^2 + (y+1)^2 = 1 \Rightarrow \text{center } C_2 = (1,-1),\ r_2 = 1$ Distance between centers: $\sqrt{(2)^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}$ Using tangent condition: \[ |-m-1+c| = 2\sqrt{m^2+1}, \quad |m+1+c| = \sqrt{m^2+1} \] Solve both equations to find possible $m$. Eliminating $c$ gives: \[ 3m^2 + 8m + 3 = 0 \Rightarrow \text{Product of roots} = \frac{3}{3} = 1 \]