JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
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
- A \(\frac{3 \sqrt{3}}{4}\)
- B \(2 \sqrt{13}\)
- C \(\sqrt{13}\)
- D \(\frac{2 \sqrt{13}}{3}\)
Answer & Solution
Correct Answer
(D) \(\frac{2 \sqrt{13}}{3}\)
Step-by-step Solution
Detailed explanation
Equation of tangent at \(A (4,-11)\) on circle is \(\Rightarrow 4 x-11 y -3\left(\frac{ x +4}{2}\right)+10\left(\frac{ y -11}{2}\right)-15=0\) \(\Rightarrow 5 x -12 y -152=0 \ldots \ldots(1)\) Equation of tangent at \(B (8,-5)\) on circle is…
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