Equation of given circle is S ≡ x2 + y2 + 6x + 8y – 11 = 0
Let L ≡ (1, –1)
For point L(1, –1), S1 = 12 + (–1)2 + 6.1 + 8(–1) – 11 = –11 and
T ≡ x.1 + y (–1) + 3(x + 1) + 4(y – 1) – 11
i.e. T ≡ 4x + 3y – 12
Now equation of the chord of circle (i) whose middle point is L(1, –1) is
T = S1 or 4x + 3y – 12 = –11 or 4x + 3y – 1 = 0
Second Method :
Let C be the centre of the given circle, thenC ≡ (–3, –4). L ≡ (1, –1) slope of CL
∴ Equation of chord of circle whose middle point is L is
[∵ chord is perpendicular to CL)
or 4x + 3y – 1 = 0