Equation of tangents from external point to a circle I have point $(p,q)$ and circle $ x^2 + y^2 + 2gx + 2fy + c = 0$.
I'm aware you could do substitute in $y = mx+c$ and solve a quadratic or you could use $ y-(-f) = m(x-(-g)) \sqrt{1+m^2}$ where m is the slope of the line and $(-g, - f)$ is the coordinates of the centre. But are there other ways? perhaps using calculus/vectors/complex numbers
maybe take point $(4,-5)$ and circle $ x^2 + y^2 -6x + -4y + 4 = 0$. as example
 A: $x^2 + y^2 -6x  +4y + 4 = 0$
centre $C(3;\;-2)$ radius $r=3$
$P(4;\;-5)$
Write the equation of the generic line passing through $P$
$y+5=m(x-4)$
adjust it in normal form
$\mathscr{F}:mx -y -4m-5=0$
Calculate the distance $d(m)$ from $\mathscr{F}$ to the centre $C$ of the circle
$$d(m)=\frac{|3m+2-4m-5|}{\sqrt{m^2+1}}$$
The line $\mathscr{F}$ is tangent to the circle if the distance $d(m)$ is equal to the radius
$$d(m)=3\to \frac{|-3-m|}{\sqrt{m^2+1}}=3$$
square both sides
$$9+6m+m^2=9m^2+9\to m_1=0;\;m_2=\frac34$$
the two tangents are 
$y=-5;\;y=\frac{3}{4}x-8$
Second method
Works with any conic
Find the equation of the polar line of $P$ wrt the circle
Substitute
$$x^2\to x_px;\;y^2\to y_py;\;x\to\frac{x+x_p}{2};\;y\to \frac{y+y_p}{2}$$
$$4x-5y-6\,\frac{x+4}{2}+4\,\frac{y-5}{2}+4=0$$
$p:x-3y=18$
The intersection points of this line with the circle are the tangent points of the tangents from $P$ to the circle, namely $K(3,-5);\;H(4.8,-4.4)$
Hope this helps
$$...$$

A: By Joachimsthal $$s_1^2=s \cdot s_{11}$$ defines the tangents from $(p,q)$ as a line pair with  $$s=x^2 + y^2 + 2gx + 2fy + c = 0,$$ $$s_{11}=p^2 + q^2 + 2gp + 2fq + c,$$ $$s_1=xp + yq + g(x+p) + f(y+q) + c.$$
I.e.
$(xp + yq + g(x+p) + f(y+q) + c)^2-(x^2 + y^2 + 2gx + 2fy + c)(p^2 + q^2 + 2gp + 2fq + c)=0$
or
${\small (-g^2+2fq+q^2+c)(x-p)^2+(-2fg-2fp-2gq-2pq)(x-p)(y-q)+(-f^2+2gp+p^2+c)(y-q)^2=0}$
which factorises as:
$${\Tiny \frac{((-g^2+2fq+q^2+c)(x-p) -(y-q) (pq+gq+fp+fg-\sqrt{D}))
((-g^2+2fq+q^2+c)(x-p)-(y-q)(pq+gq+fp+fg+\sqrt{D}))}{-g^2+2fq+q^2+c}},$$ where $$D=(g^2+f^2-c)(p^2+q^2+2gp+2fq+c)=(g^2+f^2-c)s_{11},$$
or
$${\Tiny \frac{((p^2+2gp-f^2+c)(y-q) -(x-p) (pq+gq+fp+fg-\sqrt{D}))
((p^2+2gp-f^2+c)(y-q)-(x-p)(pq+gq+fp+fg+\sqrt{D}))}{p^2+2gp-f^2+c}},$$
making the actual lines
$$(y-q)=(x-p)\frac{pq+gq+fp+fg\pm \sqrt{D}}{p^2+2gp-f^2+c}.$$
