JEE Main 2023 · 29 Jan · Shift 1

JEE Main 2023 Mathematics question on Circles

Let the tangents at the points A(4,−11)A(4,-11) and B(8,−5)B(8,-5) on the circle x2+y2−3x+10y−15=0x^2 + y^2 - 3x + 10y - 15 = 0, intersect at the point CC. Then the radius of the circle, whose centre is CC and the line joining AA and BB is its tangent, is equal to
  1. A.
    334\dfrac{3\sqrt{3}}{4}
  2. B.
    13\sqrt{13}
  3. C.
    2133\dfrac{2\sqrt{13}}{3}
  4. D.
    2132\sqrt{13}
Show answer

Correct option: C

[JEE Main 2023 · 29 Jan · Shift 1]

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