Question

The equation of the circle whose diameter is the common chord of the circles x² + y² + 2x + 3y + 2 = 0 and

x² + y² + 2x –3y – 4 = 0 is-

• A. x² + y² + 2x + 2y + 2 = 0
• B. x² + y² + 2x + 2y – 1 = 0
• C. x² + y² + 2x + 2y + 1 = 0
• D. x² + y² + 2x + 2y + 3 = 0

Answer 1

Answer: (C)

x² + y² + 2x + 2y + 1 = 0

Answer 2

The equation of common chord is S – SΆ = 0 is

6y + 6 = 0 ή y + 1 = 0

The equation of circles passing through the intersection of given circle is

x² + y² + 2x + 3y + 2 +l (y +1) = 0

x² + y² + 2x + y (3 +l) + 2 + l = 0

centre $\left( - 1, - \frac{3 + \lambda}{2} \right)$

line on y + 1 = 0

– $\left( \frac{3 + \lambda}{2} \right)$+ 1 = 0

–3 – l + 2 = 0

l = –1

required equation of circle is x² + y² + 2x + 2y + 1 = 0