Finding the equation of the line tangent to $x^2+y^2-6x+4y-3=0$ passing through $(7,4)$. How to proceed, not knowing point of tangency? 
Find the equation of the tangent line to the circle $x^2+y^2-6x+4y-3=0$ which passes through the point $(7,4)$.

Graphing the circle, $(7,4)$ is a point not on the circle. So, I am assuming it's on the tangent line.

How do you find the equation of the tangent line? Explanation needed.


Without calculus, refer to the graph:
$\hspace{2cm}$
To find the line $AB$, let $B(x,y)$, then:
$$\begin{cases}AB=AC=6 \\ OB=4\end{cases} \Rightarrow \begin{cases} (7-x)^2+(4-y)^2=6^2 \\ (3-x)^2+(-2-y)^2=4^2 \end{cases} \Rightarrow \\
(x,y)=C(7,-2); B\left(\frac{19}{13}, \frac{22}{13}\right)$$
Hence, the tangent line $AC$ is $x=7$ and the tangent line $AB$ is $y=\frac5{12}x+\frac{13}{12}.$
