<s>[INST] 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}
 [/INST] 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
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