Let the tangents at the points A ( 4, - 11 ) and B ( 8, - 5 ) on...

Given,

The tangents at the points $A ( 4, - 11 )$ and B ( 8, - 5 )$on the circle${x}^{2} + {y}^{2} - 3x + 10y - 15 = 0$, intersect at the point$C$. </p><p>So, equation of tangent at$A ( 4, - 11 )$on circle will be</p><p>$⇒ 4x - 11y - 3\left(\dfrac{x + 4}{2}\right) + 10\left(\dfrac{y - 11}{2}\right) - 15 = 0$</p><p>$⇒ 5x - 12y - 152 = 0 . . . . . . . . . \left(1\right)$</p><p>And equation of tangent at$B ( 8, - 5 )$on circle is</p><p>$⇒ 8x - 5y - 3\left(\dfrac{x + 8}{2}\right) + 10\left(\dfrac{y - 5}{2}\right) - 15 = 0$</p><p>$⇒ 13x - 104 = 0 ⇒ x = 8 . . . . . . . . . \left(2\right)$</p><p>Now putting the value of$x = 8$in equation \left(1\right)$ we get, $y = \dfrac{- 28}{3}$

So, the point $C$ will be $C\left(8, \dfrac{- 28}{3}\right)$

Now equation of tangent $AB$ will be,

$y + 11 = \dfrac{- 5 + 11}{8 - 4}\left(x - 4\right)$

$⇒ 3x - 2y - 34 = 0$

$r = \left|\dfrac{3 \times 8 + \dfrac{2 \times 28}{3} - 34}{\sqrt{13}}\right| = \left|\dfrac{- 26}{3\sqrt{13}}\right| = \dfrac{2\sqrt{13}}{3}$