Question

Find the equation of chord of contact of circle $x^2+y^2-3x+4y+5=0$ w.r.t point (1,2)

Answer 1

Equation of the circle is given as,

$x^2+y^2-3x+4y+5=0$ ………….$\left(1\right)$

$\Rightarrow x^2+y^2+2(-3/2)x+2.\left(2\right)y+5=0$

Where $g=-3/2,f=2,c=5$

$\therefore$ Equation of the chord of contact is given by

$x_{1}x+y_{1}y+g(x_1+x)+f(y_1+y)+c=0$

$\Rightarrow x_{1}x+y_{1}y+\left(\frac{-3}{2}\right)(x_1+x_2)+2(y_1+y)+5=0$

Hence equation of the chord of contact of circle $\left(1\right)$ w.r.t $(1,2)$

$1x+2y+\left(\frac{-3}{2}\right)(1+x)+2(2+y)+5=0$

$\Rightarrow x+2y-\frac{3}{2}-\frac{3x}{2}+4+4y+5=0$

$\Rightarrow \frac{2x+4y-3-3x+8+8y+10}{2}=0$

$\Rightarrow 2x+4y-3-3x+8+8y+10=0$

$\Rightarrow -x+12y+15=0$

$\Rightarrow x-12y-15=0$

$\Rightarrow x-12y=15$.