Question

Find the equation of pair of tangents drawn from the point $(0,1)$ to the circle $x^2+y^2-2x+4y=0$

Answer 1

A point $(x_1,y_1)$ is outside a circle $x^2+y^2+2gx+2fy+c=0$ if ${x_1}^2+{y_1}^2+2g{x}_1+2f{y}_1+c=0$

Given:$x^2+y^2-2x+4y=0$

The point $(0,1)$ lies $x^2+y^2-2x+4y=0+1-0+4=5>0$ outside the circle.

Hence we have two tangents

Euation of line with slope $m$ and passing through $(0,1)$ is $y-1=m(x-0)$

or $y=mx+1$

Substitute the value of $y=mx+1$ in the equation $x^2+y^2-2x+4y=0$ we get

$x^2+{(mx+1)}^2-2x+4(mx+1)=0$

$(1+m^2)x^2+(6m-2)x+5=0$ is quadratic in $x$

$\Rightarrow$ The line $y=mx+1$ would be a tangent if it cuts the circle only at one point which is true if the discriminant$=0$

$\Rightarrow {(6m-2)}^2-20(1+m^2)=0$

$\Rightarrow 36m^2+4-24m-20-20m^2=0$

$\Rightarrow 16m^1-24m-16=0$

$\Rightarrow 2m^1-3m-2=0$

$\Rightarrow (2m+1)(m-2)=0$

$\Rightarrow m=\frac{-1}{2},2$

So, equation of tangents are $y=2x+1,y=\frac{-1}{2}x+1$

So, the equation of pair of tangents are $(y-2x-1)(2y+x-2)=0$

or $2x^2-2y^2+3xy-3x+4y-2=0$