Question

Find the equation of circle whose diameter is common chord of the
circles x2 + y2 – 4x – 6y - 12 = 0 and x2 + y2 + 6x + 4y - 12 = 0. (5M)

Answer 1

Answer:

Here the common chord is the radical axis of (1) and (2). The equation of the radical axis is S – S′ = 0. i.e., 2x + 1 = 0 —— (3) The equation of any circle passing through the points of intersection of (1) and (3) is (S + λL = 0) (x2 + y2 + 2x + 3y + 1) + λ(2x + 1) = 0 x2 + y2 + 2(λ + 1)x + 3y + (1 + λ) = 0 — (4) The centre of this circle is (− (λ + 1), 3/2). For the circle (4), 2x + 1 = 0 is one chord. This chord will be a diameter of the circle (4) if the centre of (4) lies on (3). ∴ 2{–(λ + 1)} + 1 = 0 ⇒ λ = – 1/2 Thus equation of the circle whose diameter is the common chord (1) and (2) [put λ = 1/3 in equation (4)] 2(x2 + y2) + 2x + 6y + 1 = 0Read more on Sarthaks.com - https://www.sarthaks.com/535929/find-the-equation-of-the-circle-whose-diameter-is-the-common-chord-of-the-circles-x-2-y-2-2x-3y