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

Step-by-step solution

Given, 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 . So, equation of tangent at A ( 4 , - 11 ) on circle will be ⇒ 4 x - 11 y - 3 x + 4 2 + 10 y - 11 2 - 15 = 0 ⇒ 5 x - 12 y - 152 = 0   . . . . . . . . . 1 And equation of tangent at B ( 8 , - 5 ) on circle is ⇒ 8 x - 5 y - 3 x + 8 2 + 10 y - 5 2 - 15 = 0 ⇒ 13 x - 104 = 0 ⇒ x = 8   . . . . . . . . . 2 Now putting the value of x = 8 in equa