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: \[ 3m^2 + 8m + 3 = 0 \Rightarrow \text{Product of roots} = \frac{3}{3} = 1 \]

Answer 2

Step 1: Identify the given circles
Circle 1: \(x^2 + y^2 + 2x - 2y - 2 = 0\)
Circle 2: \(x^2 + y^2 - 2x + 2y + 1 = 0\)

Step 2: Find centers and radii of the circles
Complete the square for each circle:

For Circle 1:
\[ x^2 + 2x + y^2 - 2y = 2 \] \[ (x^2 + 2x + 1) + (y^2 - 2y + 1) = 2 + 1 + 1 = 4 \] \[ (x + 1)^2 + (y - 1)^2 = 4 \] Center \(C_1 = (-1, 1)\), radius \(r_1 = 2\)

For Circle 2:
\[ x^2 - 2x + y^2 + 2y = -1 \] \[ (x^2 - 2x + 1) + (y^2 + 2y + 1) = -1 + 1 + 1 = 1 \] \[ (x - 1)^2 + (y + 1)^2 = 1 \] Center \(C_2 = (1, -1)\), radius \(r_2 = 1\)

Step 3: Use formula for product of slopes of common tangents
For two circles, the product of slopes \(m_1 m_2\) of their common tangents satisfies:
\[ m_1 m_2 = \frac{c_1 c_2}{a_1 a_2} \] where \(a_i\) and \(c_i\) are coefficients from the general tangent line \(y = m x + c\) touching the circles.

Alternatively, note that for common tangents of two circles with centers \(C_1\) and \(C_2\), the product of slopes of common tangents equals 1 if the centers lie on the line \(y = -x\) or are symmetric with respect to this.

Step 4: Verify geometrically
Centers: \(C_1(-1, 1)\), \(C_2(1, -1)\) lie on the line \(y = -x\). This implies the product of slopes of the common tangents is 1.

Final answer:
\[ \boxed{1} \]