Circles
Chord of Contact — Radius of New Circle
nta_pyq_2023_jan
Grade None
Question:
Let 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 $C$. Then the radius of the circle, whose centre is $C$ and the line joining $A$ and $B$ is its tangent, is equal to:
\dfrac{3\sqrt{3}}{4}
2\sqrt{13}
\sqrt{13}
\dfrac{2\sqrt{13}}{3}
Step-by-Step Solution
Key Concept: Tangent at $A$: $4x-11y-3(x+4)/2+10(y-11)/2-15=0\Rightarrow5x-12y-152=0$. Tangent at $B$: $\Rightarrow x=8\Rightarrow y=28/3$. So $C=(8,28/3)$.
Radius $=\dfrac{2\sqrt{13}}{3}$.
Correct Answer: 4