Question

Find the equation of the chord of the circle $x^2+y^2+6x+8y-11=0$ whose middle point is (1, -1)

Answer 1

Equation of given circle is S = $x^2+y^2+6x+8y-11=0$
Let L = (1, -1) For point L(1, -1), $S_1=1_2+(-1{)}^2+6.1+8(-1)-11=-11$ and
$T\equiv x.1+y(-1)+3(x+1)+4(y-1)-11$ i.e. $T\equiv$ $4x+3y-12$
Now equation of the chord of circle (i) whose middle point is L(1, -1) is
$T=S_{1}or4x+3y-12=-11or4x+3y-1=0$
Second Method : Let C be the centre of the given circle then $C\equiv (-3,-4).L\equiv (1,-1)$
$CL=\frac{-4+1}{-3-1}=\frac{3}{4}$
$\therefore$ Equation of chord of circle whose middle point is $y+1=-\frac{4}{3}(x-1)$
($\because$ chord is perpendicular to CL) or $4x+3y-1=0$