Point Q is moving on the circle (x – 4)² + (y – 8)² = 20. Then it broke away from it and moving along a trangent to the circle, cuts x-axis at the point (–2, 0). The coordinates

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$(\frac{- 2}{5}, \frac{44}{5})$

(3, 5)

(6, 4)

$(\frac{- 3}{5}, \frac{46}{5})$

answer is B, C.

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Detailed Solution

$\sin θ = \frac{2 \sqrt{5}}{10} = \frac{1}{\sqrt{5}}$
Slope of PA, PB are $\tan (α \pm θ)$
Where $\tan α = \frac{8}{6} = \frac{4}{3} (slope of PC)$
$∴ \frac{\frac{4}{3} + \frac{1}{2}}{1 - \frac{4}{3} \cdot \frac{1}{2}}, \frac{\frac{4}{3} - \frac{1}{2}}{1 + \frac{4}{3} \cdot \frac{1}{2}} = \frac{11}{2}, \frac{5}{10}$
$\begin{array}{l} ∴ A, B = (4 + 2 \sqrt{5} (\frac{- 11}{5 \sqrt{5}}), 8 + 2 \sqrt{5} (\frac{2}{5 \sqrt{5}})) & \\ (4 - 2 \sqrt{5} (\frac{- 11}{5 \sqrt{5}}), 8 + 2 \sqrt{5} (\frac{2}{5 \sqrt{5}})) \end{array}$