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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!

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  • $\begingroup$ The Theorem of Thales is useful here. Look at the circle which has, as diameter, the line segment connecting the center of your circle and the external point. $\endgroup$ – lulu Aug 10 '15 at 3:25
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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 $$\bbox[5px, border:2px solid #C0A000]{\color{red}{y=\left(\frac{50+6\sqrt{29}}{91}\right)x+\frac{160- 90\sqrt{29}}{91}}}$$ &

$$\bbox[5px, border:2px solid #C0A000]{\color{red}{y=\left(\frac{50-6\sqrt{29}}{91}\right)x+\frac{160+90\sqrt{29}}{91}}}$$

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(1) Find X(p, q), the midpoint of C(5, 5) and P(15, 10).

(2) Find R, the length between X and C.

(3) Use X as center and R as radius to draw a circle cutting the original at M, and N.

(4) Use two-point form to get the equations of the required tangent.

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