# The point of intersection of tangents of a circle and the circumcircle formed by the points of contact and the center of the original circle.

I just wanted to know how this result is derived.

Let there be a circle whose equation is $$x^2+y^2=a^2$$. Let there be a chord PQ. If we draw the tangents from points P and Q they will intersect at a point (say, T). Now if we construct the circumcircle of the triangle OPQ ( O being the center of the initial circle), then Why is it so that T lies on the circumcircle of the triangle OPQ? I have already tested this result several times but can't figure out the derivation...

• $OPTQ$ is cyclic becuase $OP \perp PT$ and $OQ\perp QT$. Feb 16, 2021 at 15:58

The circumference centered at $$T$$ with radius $$TP$$ is orthogonal to the circumference centered at $$O$$ with radius $$OP$$.
Let me first recall this very elemental geometry Theorem: Let $$C$$ be a circumference and let $$P$$, $$Q$$ and $$R$$ be points in $$C$$ with $$P$$ and $$Q$$ diametrically opposite. Then the angle $$\angle{PRQ}=90^º.$$
With this being said, in your problem, the angle $$\angle{OPT}=90^º$$ (by the way you defined $$T$$). By the Theorem, $$O$$, $$T$$ and $$P$$ are in a circumference ($$C_1$$) where $$OT$$ is a diameter. By the same reasoning, the $$\angle{OQT}=90^º$$ and therefore $$O$$, $$T$$ and $$Q$$ are in a circumference ($$C_2$$) where $$OT$$ is a diameter. Finally, since $$C_1$$ and $$C_2$$ contain $$O$$ and $$T$$ and have the same diameter, then $$C_1$$ and $$C_2$$ are the same. Therefore $$O$$, $$P$$, $$T$$, and $$Q$$ lay on the same circumference.