Inspired by My Best Friend - 2

Consider the given circle to be $x^2 + y^2 = 400$ . Let the point P be located at a distance $d$ from the origin. Considering the right triangle formed by tangent to the circle from P, the line segment joining P to the origin, and the radius joining origin to the point of contact of the tangent from P, we get, using the Pythagoras theorem, $OP = 20\sqrt{2}$ . Similarly $OQ = 4\sqrt{61}$ . Therefore the coordinates of P and Q can be written as $( 20\sqrt{2}\cos\alpha , 20\sqrt{2}\sin\alpha )$ and $( 4\sqrt{61}\cos\beta , 4\sqrt{61}\sin\beta )$ .

Using the result that the polar of a point $(h,k)$ with respect to a circle $x^2 + y^2 = r^2$ is $xh + yk = r^2$ , substituting the point as P, and getting the condition for line passing through Q we get, $\cos( \alpha - \beta ) = \frac{5}{\sqrt{122}}$ .

In this way, in triangle OPQ, we have found lengths OP and OQ and angle between OP and OQ. The distance PQ can now be obtained by using cosine rule.