Tangent of a circle While I was preparing for $AIME$, I ran into a question regarding circles and their tangents(on a coordinate plane). I've read several posts regarding similar question types but I haven't found a systematic way of finding tangent points on a circle given the equation of a circle and a point outside the circle that is intersected by the tangent/s. 
For example, 
Find the equation of the tangent/s of circle a with center $(5, 5)$ and a radius $3$, and intercepting the point $(15, 10)$.
Any non-calculus based methods would be greatly appreciated!

Notice, let the equation of the tangent be $y=mx+c$ then satisfying this equation by the point $(15, 10)$ , we get $$10=m(15)+c$$ $$15m+c=10$$ 
$$c=10-15m\tag 1$$ 
Now, the length of perpendicular from the center $(5, 5) $ to the tangent $y=mx+c$ must be equal to the radius $3$ of the circle, hence we have  $$\frac{|m(5)-5+c|}{\sqrt{m^2+(-1)^2}}=3$$
$$\frac{|5m-5+10-15m|}{\sqrt{m^2+1}}=3$$
$$|5-10m|=3\sqrt{m^2+1}$$
$$25+100m^2-100m=9m^2+9$$ $$91m^2-100m+16=0$$ 
On solving above quadratic equation for $m$, we get 
$$m=\frac{50\pm 6\sqrt{29}}{91}$$ By substituting these values in the eq(1), we get $$m=\frac{50+6\sqrt{29}}{91}\implies c=10-15\left(\frac{50+6\sqrt{29}}{91}\right) =\frac{160- 90\sqrt{29}}{91}$$
$$m=\frac{50-6\sqrt{29}}{91}\implies c=10-15\left(\frac{50-6\sqrt{29}}{91}\right) =\frac{160+ 90\sqrt{29}}{91}$$
Hence, we get two tangent lines from the external point $(15, 10)$ to the given circle as follows $$\bbox[5px, border:2px solid #C0A000]{\color{red}{y=\left(\frac{50+6\sqrt{29}}{91}\right)x+\frac{160- 90\sqrt{29}}{91}}}$$ &
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{y=\left(\frac{50-6\sqrt{29}}{91}\right)x+\frac{160+90\sqrt{29}}{91}}}$$ 
