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}
Best Answer
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 $cos(\theta)=2/5$