Question

The centres of all circles passing through the points of intersection of the circles \(x^2 + y^2 + 2x - 2y + 1 = 0\) and \(x^2 + y^2 - 2x + 2y - 2 = 0\) and having radius \(\sqrt{14}\) lie on the curve

• A. \(x + y = 0\)
• B. \(y^2 = 4x - 2\)
• C. \(3x^2 + 5x = y\)
• D. \(2x^2 + 3y^2 = 7\)

Hint:

For any two intersecting circles, the line of centers is the perpendicular bisector of the common chord. Any circle passing through the intersection points must have its center on this line.

Answer 1

The Correct Option is A

Step 1: Understand the Geometric Property:
The centers of any family of circles passing through the intersection points of two given circles \(S_1 = 0\) and \(S_2 = 0\) always lie on the perpendicular bisector of the common chord. The perpendicular bisector of the common chord is simply the line joining the centers of the two given circles.
Step 2: Find Centers of Given Circles:
Circle 1: \(x^2 + y^2 + 2x - 2y + 1 = 0\) Center \(C_1 = (-1, 1)\). Circle 2: \(x^2 + y^2 - 2x + 2y - 2 = 0\) Center \(C_2 = (1, -1)\).
Step 3: Find Equation of Line Joining Centers:
The locus of the centers of the required circles is the line passing through \(C_1(-1, 1)\) and \(C_2(1, -1)\). Slope \(m = \frac{-1 - 1}{1 - (-1)} = \frac{-2}{2} = -1\). Equation using point-slope form with \(C_1\): \[ y - 1 = -1(x - (-1)) \] \[ y - 1 = -(x + 1) \] \[ y - 1 = -x - 1 \] \[ x + y = 0 \]
Step 4: Conclusion:
Regardless of the radius condition (which would just select specific points on this line), the centers lie on the line \(x + y = 0\).