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Let the tangents at the points $A (4, - 11) and B (8, - 5)$ on the circle $x^{2} + y^{2} - 3 x + 10 y - 15 = 0$ , intersect at the point C. then the radius of the circle, whose centre is C and the line joining A and B is its tangent, is equal to

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a

$\frac{2 \sqrt{13}}{3}$

b

$2 \sqrt{13}$

c

$\frac{3 \sqrt{3}}{4}$

d

$\sqrt{13}$

answer is D.

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$\begin{array}{l} r = \sqrt{\frac{9}{4} + 25 + 15} = \sqrt{\frac{169}{4}} = \frac{13}{2} \\ AB = \sqrt{16 + 36} = 2 \sqrt{13} \\ AM = \sqrt{13} \\ P M = \sqrt{{(\frac{13}{2})}^{2} - 13} = \frac{\sqrt{117}}{2} \\ \tan θ = \frac{\sqrt{13}}{CM} \\ CM = \sqrt{13} \cot θ \\ = \sqrt{13} \cdot \frac{\sqrt{13}}{\sqrt{117}} \times 2 \\ = \frac{2 \sqrt{13}}{3} \end{array}$

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