How to find the center of a circle from a point on it, and a tangent line. I am clearly not typing this in to search engines correctly, if it's possible. The teacher likes to ask trick questions, so it might actually not be possible. I can't see anything like this in my notes, nor does anything come up when I Google. 
When I turned in the work, I assumed the point on the circle was opposite to the tangent line. I was told this wasn't a valid assumption to make. I'm also not getting any answers from him. I can't think of any way to solve this, so I turn to you guys.

Consider a circle $C$ that is tangent to $3x+4y-12=0$ at $(0,3)$ and contains $(2,-1)$. 
      Set up equations that would determine the center $(h,k)$ and radius $r$ from the circle. 
      DO NOT SOLVE THESE EQUATIONS.

 A: You can find the coordinates of the center as a intersection of two lines. First of all, since the points $(0,3)$ and $(2,-1)$ are in the circle, the center is equidistant to them. The locus of points equidistant from two given points is a line (called perpendicular bisector), so you have the first line containing the center there. On the other hand, you also know that the line perpendicular to the tangent line at $(0,3)$ contains the center (basic theorem from plane geometry). You can find this line because you know its slope and the coordinates of a point through which it passes. Finally, the intersection of these two lines would give you the coordinates of the center because they both contain it.
A: The gradient of the tangent at $(0,3)$ is $-3/4$. A line from the centre of the circle to $(0,3)$ therefore has gradient $4/3$. Let the centre of the circle be $(h,k)$. Then the gradient of this line is
$$\frac{k-3}{h}=\frac{4}{3} \iff 3k-9=4h$$
Since $(2,-1)$ lies on the circle,
$$(2-h)^2+(-1-k)^2=r^2.$$
Since $(0,3)$ lies on the circle, 
$$h^2+(3-k)^2=r^2.$$
Use these three equations to find $k,h,r$.
A: Let's start from the obvious, let's assume that : 
$$C : (x-x_0)^2 + (y-y_0)^2 = r^2$$
where $(x_0,y_0) = (h,k)$ the center of the circle.
Since $(2,-1) \in C$, then : 
$$(2-x_0)^2 + (-1-y_0)^2 = r^2$$
On the same time, since the point that the circle meets the tangent is $(0,3)$, it means that this point also belongs to the circle, so : 
$$(0-x_0)^2 + (3-y_0)^2 = r^2$$
Considering the equation of the tangent, we have : 
$$3x+ 4y-12 = 0 \Leftrightarrow 4y =-3x + 12 \Leftrightarrow y=-\frac{3}{4}x + 3$$
which tells us that $λ=-\frac{3}{4}$.
Now, considering a straight line that goes by the center of the cirle $(x_0,y_0)$ and through $(0,3)$, would be perpendicular to the tangent line (draw a sketch to see that), which means that : 
$$λ\cdotλ_1=-1\Rightarrow λ_1 = \frac{4}{3}$$
So, the gradient of this line should be : 
$$\frac{y_0-3}{x_0-0}=\frac{4}{3}$$
Now, solving the following system will yield your center and your radius : 
$$\begin{cases} (2-x_0)^2 + (-1-y_0)^2 = r^2 \\ (0-x_0)^2 + (3-y_0)^2 = r^2 \\ \frac{y_0-3}{x_0-0}=\frac{4}{3}
\end{cases}$$
Tip (1) : Remember that in such problems it's always good to write down the equations you know regarding the geometric shape you're studying and thus forming out systems that will lead you through.
Tip (2) : Analytic Geometry is a lovely lesson but also requires you to be thorough. Never be afraid to make a graph, never be afraid to contain all the details ! (You avoid mistakes by this way too !)
A: As I was looking for a vector based solution for this problem and could not find a satisfactory answer I decided to post my own solution here just in case someone needs it.
setup


*

*(known) tangent defined via point P0 and unit vector v0.

*(implicitly known) n0 as normal vector (v0 rotated 90°)

*(known) second point on circle P1

*(unknown) circle center C
approach
It is known that C is on the line defined by P0 and n0:
$\left(\begin{matrix}C_x\\C_y\\\end{matrix}\right)=\left(\begin{matrix}P_{0x}\\P_{0y}\\\end{matrix}\right)+r\left(\begin{matrix}n_{0x}\\n_{0y}\\\end{matrix}\right)\ (eq1)$
Secondly we know that the distances between both points P0, P1, and the center C must be equal:
$\left|\bar{P_0C}\right|=\left|\bar{P_1C}\right|\ $
$\left(C_x-P_{0x}\right)^2+\left(C_y-P_{0y}\right)^2=\left(C_x-P_{1x}\right)^2+\left(C_y-P_{1y}\right)^2\ (eq2)$
Now we can put eq1 into eq2:
$\left(P_{0x}+r{\ n}_{0x}-P_{0x}\right)^2+\left(P_{0y}+r{\ n}_{0y}-P_{0y}\right)^2=\left(P_{0x}+r{\ n}_{0x}-P_{1x}\right)^2+\left(P_{0y}+r{\ n}_{0y}-P_{1y}\right)^2$
which can be simplified to:
$\left(r{\ n}_{0x}\right)^2+\left(r{\ n}_{0y}\right)^2=\left((P_{0x}-P_{1x})+r{\ n}_{0x}\right)^2+\left((P_{0y}-P_{1y})+r{\ n}_{0y}\right)^2$
with $\Delta P=\left(\begin{matrix}\Delta_x\\\Delta_y\\\end{matrix}\right)=\left(\begin{matrix}P_{0x}-P_{1x}\\P_{0y}-P_{1y}\\\end{matrix}\right)$ it follows:
$\left(r{\ n}_{0x}\right)^2+\left(r{\ n}_{0y}\right)^2=\left(\Delta_x+r{\ n}_{0x}\right)^2+\left(\Delta_y+r{\ n}_{0y}\right)^2$
breaking up the brackets and reordering for r leads to:
$r^2({{\ n}_{0x}}^2+{{\ n}_{0y}}^2-{{\ n}_{0x}}^2-{{\ n}_{0y}}^2)+2r(\Delta_x{\ n}_{0x}+\Delta_y{\ n}_{0y})+({\Delta_x}^2+{\Delta_y}^2)=0$
The $r^2$ component eliminates itself and I can solve for r:
solution
$r=\frac{-({\Delta_x}^2+{\Delta_y}^2)}{2(\Delta_x{\ n}_{0x}+\Delta_y{\ n}_{0y})}$
the center C therefore calculates as:
$C\ =P_0\ +\ r\ n_0$
It is important to remember that r is signed, indicating the side of the tangent that the circle is on.
A: The equation of the tangent line is given as $ \ 3x + 4y \ = \ 12 \ \ , \ $ so we may find the normal line at $ \ (0 \ , \ 3) \ $ by "swapping" the coefficients in the linear equation and changing the sign of one of them to produce $ \ 4x \ - \ 3y \ = \ c \ \ ; \ $ inserting the coordinates of the tangent point then gives us $ \ 4·0 \ - \ 3·3 \ = \ -9 \ = \ c \ \ . \ $  So the center of the circle lies somewhere on this line, which we can describe parametrically by $ \ x \ = \ 3t \ \ , \ \ y \ \ = \ \ 4t + 3 \ \ . \ $
The distance-squared from the center of the circle (hence, the radius-squared for the circle) to each of the given points must be equal, so we may write
$$ ( \ 3t \ - \ 0 \ )^2 \ + \ ( \ [4t + 3] - 3 \ )^2 \ \ = \ \ ( \ 3t \ - \ 2 \ )^2 \ + \ ( \ [4t + 3] - [-1] \ )^2 $$
$$ \Rightarrow \ \ 9t^2 \ + \ 16t^2 \ \ = \ \ ( \ 9t^2 \ - \ 12t \ + \ 4 \ )  \ + \ ( \ 16t^2 \ + \ 32t \ + \ 16 \ ) $$
$$ \Rightarrow \ \ 20t \ + \ 20 \ \ = \ \ 0 \ \ \Rightarrow \ \ t \ = \ -1 \ \ . \ \ $$
Since this proves to be a "disguised" linear equation for $ \ t \ \ , \ $ the solution for the circle is unique.  We conclude that the center of the circle is located at $ \ ( \ 3·[-1] \ , \ 4·[-1] + 3 \ ) \ = \ (-3 \ , \ -1) \ \ $ and the radius-squared is $ \ r^2 \ = \ 9t^2 + 16t^2 \ = \ 25·(-1)^2 \ = \ 25 \ \ , \ $ or a radius of $ \ 5 \ \ . \ $  (A check of the right side of the radius-squared equation also gives us $ \ ( \ 9 \ + \ 12 \ + \ 4 \ )  \ + \ ( \ 16 \ - \ 32 \ + \ 16 \ ) \ = \ 25 \ \ . \ )  $
[I guess the equations have been "set up" and solved...]

Just to illustrate Farley Oliveira's perpendicular-bisector method, the line containing the two  given points on the circle has slope $ \ \frac{3 \ - \ [-1]}{0 \ - \ 2} \ = \ -2 \ \ . \ $  The midpoint of the segment between the points is $ \ \left( \ \frac{0 \ + \ 2}{2} \ , \ \frac{3 \ + \ [-1]}{2} \ \right) \ = \ (1 \ , \ 1) \ \ . \ $  The equation of the perpendicular bisector of the line segment is thus $ \ y \ - \ 1 \ = \ \frac{(x \ - \ 1)}{2} \ \ . \ $  This line intersects the normal line we determined above at the point found from
$$ 4x \ - \ 3y \ = \ -9 \ \ \Rightarrow \ \ 4x \ - \ 3·\left[ \ 1 + \frac{(x \ - \ 1)}{2} \ \right] \ = \ -9 $$
$$ \Rightarrow  \ \ 8x \ - \ 6 \ - \ 3x  \ + \ 3 \ \ = \ \ -18 \ \ \Rightarrow \ \ x \ = \ -3 \ \ , \ \ y \ = \ 1 + \frac{([-3] \ - \ 1)}{2} \ = \ -1 \ \ . \ $$
So the center is found to lie at $ \ (-3 \ , \ -1) \ $ and the distance to either of the given points on the circle is $ \ 5 \ \ . $

