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 t

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Solution

Q.
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 [2023]

1 $2 \sqrt{13}$

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

3 $\sqrt{13}$

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

Ans.
(2)

$Equation of tangent at point A (4, - 11) is$

$4 x - 11 y - \frac{3}{2} (x + 4) + 5 (y - 11) - 15 = 0$

$5 x - 12 y = 152$ ...(i)

$Equation of tangent at point B (8, - 5) is$

$8 x - 5 y - \frac{3}{2} (x + 8) + 5 (y - 5) - 15 = 0$

$\Rightarrow x = 8$ ...(ii)

Solving (i) and (ii) we get

$x = 8 and y = \frac{- 28}{3}$

So, point $C$ is $(8, - \frac{28}{3})$ and this point is centre of circle whose tangent is line $A B$ .

Equation of line $A B$ is, $3 x - 2 y = 34$

Perpendicular distance of point $C$ from line $A B$ gives the radius of another circle whose centre is $C (8, - \frac{28}{3}) .$

So, $\frac{| 3 \times 8 - 2 (\frac{- 28}{3}) - 34 |}{\sqrt{3^{2} + {(- 2)}^{2}}} = \frac{2 \sqrt{13}}{3}$

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