MathematicsHard127×since 2002Q3033Let the tangents at the points A(4,−11)A(4,-11)A(4,−11) and B(8,−5)B(8,-5)B(8,−5) on the circle x2+y2−3x+10y−15=0x^{2}+y^{2}-3 x+10 y-15=0x2+y2−3x+10y−15=0, intersect at the point CCC. Then the radius of the circle, whose centre is CCC and the line joining AAA and BBB is its tangent, is equal to :A2133\frac{2\sqrt{13}}{3}3213B334\frac{3\sqrt{3}}{4}433C13\sqrt{13}13D2132\sqrt{13}213Check answerSkip