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A point $P$ is given on the circumference of a circle of radius $r$.Chords $QR$ are drawn parallel to the tangent at $P$.Then how can we determine the largest possible area of triangle $PQR$?


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Here is one way.

We may assume, without loss of generality, that the circle is centered at the origin. Let $P$ be the point $(-r,0)$. Then $QR$ will be vertical. Let $\theta$ be the angle made by the line through $Q$ and the origin, and the positive $x$-axis. Then, $Q$ has coordinates $(r \cos \theta, r \sin \theta)$.

Hence, the area of the triangle can be expressed as $$ \mbox{area} = \frac{1}{2}(r \cos \theta + r) 2r \sin \theta = r^2 \sin \theta (1+\cos \theta).$$

You can then use calculus to maximize this function of $\theta$, keeping in mind that $0 \le \theta \le \pi$.

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