Calculate Tangent Points to Circle Problem
Given a circle with radius $r = 2$ and center $C = (4,2)$ and a point $P = (-1,2)$ outside the circle.
How do I calculate the coordinates of two tangent points to the circle, given that the tangents both have to go through $P$?
My (sad) try
All I can think of is finding the equation for the circle, which is
\begin{equation}
  (x-4)^{2} + (y-2)^{2} = 4.
\end{equation}
I have no idea what to do next. (I don't even if finding the circle's equation is relevant.)
Update
After using Dhanush Krishna'a answer, I can (easily) find the two intersection points:
\begin{equation}
  (x_{1,2}, y_{1,2}) = \frac{2}{5}(8, 5\pm\sqrt{21}).
\end{equation}
 A: Take the equation of the tangent to be $(y-2)=m(x+1)$
This touches the circle. Therefore the distance of this line from the centre of the circle is equal to the radius of the circle.
$${|5m|\over {1+m^2}} {=2}$$
Square and rearrange. $|m|={4\over {\sqrt 21}}$
Now you know the line. Find the point of intersection of this line with the circle.
Another solution uses the graph of these curves. The point $P$ lies on the horizontal line joining the centre of the circle. You know the distance between the point and the centre. It is $5$. You know the radius ($2$). The angle between the radius and the line joining the centre and the point P can be found out. Use this data and  the centre of the circle to find the coordinates. 
The upper coordinate is $(4-2cos(\theta),2+2sin(\theta))$ where $\theta$ is the angle that I have mentioned above. By the way $cos(\theta)=2/5$
A: There is another way to find the that may strike you as funny. Use homogeneous coordinates for the points in the place so the exterior point is located at $P = (-1,2,1)$.
The homogeneous coordinates of a circle located at $(c_x,c_y)$ with radius $r$ are
$$ C = \begin{vmatrix} 1 & 0 & -c_x \\ 0 & 1 & -c_y \\ -c_x & -c_y & c_x^2+c_y^2-r^2 \end{vmatrix} $$
such that the expression $$\begin{pmatrix} x & y & 1 \end{pmatrix} \begin{vmatrix} 1 & 0 & -c_x \\ 0 & 1 & -c_y \\ -c_x & -c_y & c_x^2+c_y^2-r^2 \end{vmatrix} \begin{pmatrix} x \\ y \\ 1 \end{pmatrix} = 0 $$ simplifies to the standard equation for a circle
$$ (x-c_x)^2 + (y-c_y)^2 - r^2 = 0 $$
Ok, back to the problem. If you take the product $L=C P$, then $L$ is a line with coordinates $L=[a,b,c]$ such as $a x+b y  + c =0$ is the equation of the line. This line connects the two tangent points.
In your case
$$ C = \begin{vmatrix} 1 & 0 & -4 \\ 0 & 1 & -2 \\ -4 & -2 & 16 \end{vmatrix} $$
so
$$ L = \begin{vmatrix} 1 & 0 & -4 \\ 0 & 1 & -2 \\ -4 & -2 & 16 \end{vmatrix} \begin{pmatrix} -1 \\ 2 \\1 \end{pmatrix} = \begin{bmatrix} -5 \\ 0 \\ 16 \end{bmatrix} $$
or $$ \left. (x,y,1) \cdot [-5,0,16] = 0 \right\} \left. 16-5 x =0 \right\} x = \frac{16}{5} $$
Plug the line equation in the circle equation to get the other coordinate
$$ \left. \left( \frac{16}{5} \right)^2 - 8 \left( \frac{16}{5} \right) + y^2 -4 y + 16 =0 \right\} y =2 \pm \frac{2\sqrt{21}}{5} $$
The coordinates are thus
$$ \begin{pmatrix} \frac{16}{5} \\ 2 - \frac{2\sqrt{21}}{5} \end{pmatrix} \mbox{  or  } \begin{pmatrix} \frac{16}{5} \\ 2 + \frac{2\sqrt{21}}{5} \end{pmatrix}$$
NOTE: The relationship between the point $P$ and the line $L$ is that of pole and polar
A: Hint: Simple geometry tells us that the distance of those points from $P$ would be $\sqrt{5^2-2^2}=\sqrt{21}$ units. So, just build a circle with radius of that length from with $P$ at the centre and calculate the intersection points of the two circles. And you're done. 
