Question

Sol. [Key Concept: From a point on the circle, the path along a tangent: use the tangent length and the perpendicular from the centre.]

$$sin\theta = \frac{2\sqrt{5}}{10} = \frac{1}{\sqrt{5}} $$Slopes of $PA$ and $PB$ are $tan(\alpha \pm \theta)$ where $tan\alpha = \frac{8}{6} = \frac{4}{3}$

$$\begin{matrix} & \ = \frac{\frac{4}{3} + \frac{1}{2}}{1 - \frac{4}{3} \cdot \frac{1}{2}},\frac{\frac{4}{3} - \frac{1}{2}}{1 + \frac{4}{3} \cdot \frac{1}{2}} = \frac{11}{2},\frac{5}{10} \\ & \ \end{matrix}$$

$\therefore A,B \equiv \left( 4 + 2\sqrt{5}\left( \frac{- 11}{5\sqrt{5}} \right),8 + 2\sqrt{5}\left( \frac{2}{5\sqrt{5}} \right) \right)$, $\left( 4 - 2\sqrt{5}\left( \frac{- 1}{\sqrt{5}} \right),8 - 2\sqrt{5}\left( \frac{2}{\sqrt{5}} \right) \right) \equiv \left( \frac{- 2}{5},\frac{44}{5} \right) = (6,4)$