Question

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

Answer 1

Any line through the point $(8,1)$ is
$y-1=m(x-8)$
$mx-y+(1-8m)=0.....\left(1\right)$

If it is a tangent then perpendicular from centre $(1,2)$ is equal to radius $\sqrt{1+4+20}=5$

$\therefore \frac{m-2+(1-8m)}{\sqrt{(m^2+1)}}=5$

$(-7m-1{)}^2=25(m^2+1)$

$49m^2+14m+1=25m^2+25$

$24m^2+14m-24=0$

$12m^2+7m-12=0$

$12m^2+16m-9m-12=0$

$(3m+4)(4m-3)=0$

$\therefore m=-\frac{4}{3},\frac{3}{4}$.

Putting the values of $m$ in $\left(1\right)$, the required tangents are

$4x+3y-35=0$ and $3x-4y-20=0$.

$m=2,-\frac{1}{2},2x-y+1=0,x+2y-2=0$.