Question

Pair of tangents are drawn from point (1,3) to circle $x^2+y^2-3x-x+3y-1=0$ Find line joining their point of contact

Answer 1

Equation of the circle is given as,

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

$\Rightarrow x^2+y^2-4x+3y-1=0$

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

$\therefore g=-2$, $f=\frac{3}{2}$ and $c=-1$

Equation of the chord of contact of circle $\left(1\right)$ is given as

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

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

Now, equation of the chord of contact of circle $\left(1\right)$ wrt to $(1,3)$

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

$2x+6y-4(1+x)+3(3+y)-2=0$

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$2$

$\Rightarrow 2x+6y-4-4x+9+3y-2=0$

$\Rightarrow -2x+9y+3=0$

$\Rightarrow 2x-9y-3=0$

$\Rightarrow 2x-9y=3$.