Question

Find the equation of the tangents to the circle.
$x^2+y^2-4x+6y-12=0$ and parallel $x+y-8=0$.

Answer 1

We have,

${\left(x+2\right)}^2+{\left(y+3\right)}^2=25={\left(5\right)}^2$

Let the equation of the tangents be $x+y+c=0$

$\begin{array}{c}\frac{\left|2-3+c\right|}{\sqrt{2}}=5\\ \frac{\left|c-1\right|}{\sqrt{2}}=5\\ c=1+5\sqrt{2}\end{array}$

$\therefore$ equation are $x+y+1+5\sqrt{2}=0$

and $x+y+1-5\sqrt{2}=0$