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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...

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    $\begingroup$ $OPTQ$ is cyclic becuase $OP \perp PT$ and $OQ\perp QT$. $\endgroup$
    – player3236
    Feb 16, 2021 at 15:58

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Short answer:

The circumference centered at $T$ with radius $TP$ is orthogonal to the circumference centered at $O$ with radius $OP$.

Long answer:

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.

Note: This is very related to the concept of ''power of a point resect to a circumference''. Check https://en.wikipedia.org/wiki/Power_of_a_point for some interesting information about it.

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