Tangent of a circle through a point not on circle

Question

Given the equation of a circle: $(x+4)^2+(y-2)^2 = 1$ find the tangent that goes through the point $(-3,3)$…

…but this point IS NOT on this circle. Unfortunately, this is not something that I have seen before and the normal formula doesn't work.

Thank you

Answer 1

A little graphing won't hurt: By inspection, the tangent lines are $x=-3$ and $y=3$.

Answer 2

You should have a system of equation the first is the circle and the second is the general equation for a line that pass from $(3,3)$ substitute one into the others and apply the tangent condition $b^2-4ac=0$ you will find the two points. Another methods is using the derivation.

Answer 3

Well there's a formula you could use:

$$(y-b)=m(x-a)\pm r\sqrt{1+m^2}$$

for a circle $(x-a)^2+(y-b)^2=r^2$

Although (3,3) ain't giving you a tangent