Let P and Q be two distinct points on a circle which has center at...

Given the circle has centre $\left(2, 3\right)$ and it passes through the point \left(0, 0\right) .$</p><p>So, the radius of the circle is$\sqrt{{2}^{2} + {3}^{2}} = \sqrt{13}$</p><p>The equation of the line which passes through the point \left(0, 0\right) \text{ and }$\left(2, 3\right)$is$y - 0 = \dfrac{3 - 0}{2 - 0}\left(x - 0\right) ⇒ y = \dfrac{3}{2}x$</p><p>So, the slope of$OC = \dfrac{3}{2}$</p><p>From the diagram, the line$PQ$is perpendicular to the line$OC .$</p><p>So, the slope of$PQ$is$- \dfrac{2}{3}$</p><p>So,$tan\theta = - \dfrac{2}{3}$</p><p>Using symmetric from of line</p><p>The coordinate of$P = \left(2 + \sqrt{13}cos\theta, 3 + \sqrt{13}sin\theta\right)$</p><p>$= \left(2 + \sqrt{13} \cdot \left(- \dfrac{3}{\sqrt{13}}\right), 3 + \sqrt{13}\left(\dfrac{2}{\sqrt{13}}\right)\right)$</p><p>$= \left(- 1, 5\right)$</p><p>The coordinate of$Q = \left(2 - \sqrt{13}cos\theta, 3 - \sqrt{13}sin\theta\right)$</p><p>$= \left(2 - \sqrt{13} \cdot \left(- \dfrac{3}{\sqrt{13}}\right), 3 - \sqrt{13}\left(\dfrac{2}{\sqrt{13}}\right)\right)$</p><p>$= \left(5, 1\right)