how to find the equation of a tangent line to a circle, given its slope and the eq. of the circle? So I have a circle: $(x-2)^2 + (y-2)^2 = 25$ and I have a tangent line to this circle, with a slope of $m= -3/4$. 
I have to find the equation of the tangent line, so I know the radius of the circle is $r = 5$ and I wrote the equation of the tangent line as:
$$y = -3/4x + h$$
So now I have to find $y, x$ and $h$, but I don't know if I can just replace $x$ and $y$ with the center points? Or do I have to find the point-line distance (and why?) 

Here is the derivation for the equation of the line with slope $m$ that is tangent to a circle of radius $r$ centered on the origin. (If the center of the circle was not centered on the origin, we can simply shift the equation we found.)
Let the required tangent line be $y=mx+b$ and the point of tangency be $(x_0, y_0)$. We need to find $b$ in terms of $m$ and $r$.
The slope of the tangent line $m$ must equal to the slope of the circle at $(x_0, y_0)$. So let's find the equation for slope of the circle at $(x_0, y_0)$.
The equation of the circle is
$$x^2 + y^2 = r^2.$$
Implicitly differentiating, we have
$$2x + 2y \frac{dy}{dx} = 0.$$
Rearranging we have
$$\frac{dy}{dx} = -\frac{x}{y}.$$
So we have $$m = -\frac{x_0}{y_0}.$$
Therefore, we have these three equations:

*

*$y_0 = m x_0 + b$

*$m = -\frac{x_0}{y_0}$

*${x_0}^2 + {y_0}^2 = r^2$
We need to find $b$ in terms of $m$ and $r$.
Solving the second equation for $x_0$ we have $x_0 = -m y_0$. Substituting this to the first equation we have
$$b = y_0 - m x_0 = y_0 - m (-m y_0) = y_0 (1+m)^2.$$
Substituting $x_0 = -m y_0$ to the third equation we have
$$m^2 {y_0}^2 + {y_0}^2 = r^2.$$
Solving for ${y_0}^2$ we have
$$y_0^2 = \frac{r^2}{m^2+1}.$$
Taking the square root we get
$$y_0 = \pm \frac{r}{\sqrt{m^2+1}}.$$
Therefore
$$b = \pm \frac{r}{\sqrt{m^2+1}} \cdot (1+m^2) = \pm r \sqrt{m^2+1}.$$
Note that we have two values for $b$, so there are actually two tangent lines with the same slope.
Hence, the tangent lines with slope $m$ for a circle of radius $r$ centered on the origin are given by
$$y = mx \pm r \sqrt{m^2+1}.$$
Finally, if the center of the circle was instead $(h, k)$, the equation of the tangent lines are
$$y-k = m(x-h) \pm r \sqrt{m^2+1}.$$
In your case, $(h, k) = (2,2)$, $r=5$, and $m = -3/4$. So the tangent lines are
$$y-2 = -\frac{3}{4}(x-2) \pm 5\sqrt{\left(-\frac{3}{4}\right)^2 + 1}.$$
Which can be simplified to
$$y = -\frac{3}{4} x + \frac{39}{4}$$
and
$$y = -\frac{3}{4} x - \frac{11}{4}.$$
