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.

 A: $$(x-3)^2+(y+2)^2=4^2$$
If $\dfrac{y-4}{x-7}=m\iff mx-y+4-7m=0$ be the tangent of the circle, the perpendicular distance of the line from the center $(3,2)$ will be $=4$(radius)
$$4=\left|\dfrac{3m-(-2)+4-7m}{\sqrt{m^2+1}}\right|$$
$$\iff16(m^2+1)=(2-4m)^2\iff m=\infty,?$$
A: HINT
The desired tangent line $L$ must pass through some point $(x,y)$ on the circle, satisfying its equation, as well as through the point $(7,4)$. The last condition implies the equation of $L$ is $$y = m(x-7)+4.$$
Implicitly differentiating the equation of the circle to find the relevant tangent line's slope yields
$$
2x + 2yy' - 6 + 4y' = 0 \implies y' = \frac{6-2x}{2y+4}
 = \frac{3-x}{y+2}
$$
So $L$ passes through some $(x^*,y^*)$ satisfying the equation of the curve, and the equation of the line itself must be
$$
y = \frac{3-x^*}{y^*+2}(x-7)+4,
$$
and the point $(x^*,y^*)$ must satisfy this as well...
A: 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 $y=\frac5{12}x+\frac{13}{12}.$
