Question

Find the equation of the tangents to the circle $x^2+y^2-2x-4y-20=0$ which pass through the point $(8,1)$.

Answer 1

Any line through the point $(8,1)$ is
$y-1=m(x-8)$
or $mx-y+(1-8m)=0.....\left(1\right)$
If it is a tangent then perpendicular from centre $(1,2)$ is equal to radius $\sqrt{1+4+20}=5$
$\therefore \frac{m-2+(1-8m)}{\sqrt{(m^2+1)}}=5$
or $(-7m-1{)}^2=25(m^2+1)$
or $49m^2+14m+1=25m^2+25$
or $24m^2+14m-24=0$
or $12m^2+7m-12=0$
or $12m^2+16m-9m-12=0$
$(3m+4)(4m-3)=0$
$\therefore m=-4/3,3/4$.
Putting the values of $m$ in $\left(1\right)$, the required tangents are
$4x+3y-35=0$
and $3x-4y-20=0$.