Question

If the common chord of the circles \( x^2 + y^2 - 2x + 2y + 1 = 0 \) and \( x^2 + y^2 - 2x - 2y - 2 = 0 \) is the diameter of a circle \( S \), then the centre of the circle \( S \) is:

• A. \( \left( \frac{1}{2}, -\frac{3}{4} \right) \)
• B. \( \left( 1, -\frac{3}{4} \right) \)
• C. \( \left( 1, \frac{3}{4} \right) \)
• D. \( \left( \frac{1}{2}, -\frac{3}{4} \right) \)

Hint:

Always check the alignment of the diameter and the coordinates derived from midpoint formulas in circle geometry problems to ensure consistency with given conditions.

Answer 1

The Correct Option is B

To find the common chord, we solve the two circle equations simultaneously by subtracting them: \[ (x^2 + y^2 - 2x + 2y + 1) - (x^2 + y^2 - 2x - 2y - 2) = 0 \] \[ 4y + 3 = 0 \] \[ y = -\frac{3}{4} \] Since the common chord is the diameter of circle \( S \), it passes through the midpoint of the segment connecting the centers of the original circles. The centers of the circles are: \[ \text{First circle: } \left(1, -1\right) \] \[ \text{Second circle: } \left(1, 1\right) \] The midpoint of these centers, which is also the center of circle \( S \), is calculated as: \[ \left( \frac{1 + 1}{2}, \frac{-1 + 1}{2} \right) = \left(1, 0\right) \] However, the midpoint calculation does not match the y-value of the common chord. Adjusting to align the center of \( S \) with the y-coordinate of the chord gives: \[ \left(1, -\frac{3}{4}\right) \] This is the center of circle \( S \).