Question

The equation of the pair of transverse common tangents drawn to the circles \( x^2 + y^2 + 2x + 2y + 1 = 0 \) and \( x^2 + y^2 - 2x - 2y + 1 = 0 \) is:

• A. \(x^2 - y^2 = 0\)
• B. \(xy = 0\)
• C. \(x^2 - y^2 + 2x + 1 = 0\)
• D. \(x^2 - y^2 - 2y - 1 = 0\)

Hint:

To find the equation of the common tangents to two circles, use the relationship between their centers and tangents.
- The transverse common tangent satisfies the equation \( xy = 0 \) when the centers lie symmetrically.

Answer 1

The Correct Option is B

Step 1: The general form of the equation of tangents to the circles.
We have two circles: \[ x^2 + y^2 + 2x + 2y + 1 = 0 \quad \text{(Circle 1)} \] \[ x^2 + y^2 - 2x - 2y + 1 = 0 \quad \text{(Circle 2)}. \] For the transverse common tangents of these two circles, the equation can be given as: \[ (x_1 x_2 - y_1 y_2) = 0, \] where the centers of the circles are \((h_1, k_1)\) and \((h_2, k_2)\). After simplifying, we get: \[ xy = 0. \] Thus, the equation of the transverse common tangents is \( xy = 0 \), which corresponds to option (2). Hence, the final answer is \(\boxed{xy = 0}\).