Question

Find the equation of the chord of the circle x² + y² + 6x + 8y – 11 = 0, whose middle point is (1, –1) 

Answer 1

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