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JEE Main 201815 AprilEvening ShiftOnline
Exam
JEE Main
Year
2018
Date
15 April
Shift
Evening Shift
Mode
Online
Subject
Mathematics
Chapter
Circle

JEE Main 2018 · 15 April · Evening Shift · Online · Mathematics · Circle

The tangent to the circle C_1: x^2+y^2-2 x-1=0 at the point (2,1)...

The tangent to the circle C1:x2+y2−2x−1=0C_1: x^2+y^2-2 x-1=0 at the point (2,1)(2,1) cuts off a chord of length 4 from a circle C2C_2 whose centre is (3,−2)(3,-2). The radius of C2C_2 is

Options

  1. A6\sqrt{6}
  2. B2
  3. C2\sqrt{2}
  4. D3

Correct answer

A. 6\sqrt{6}

Step-by-step solution

Here, equation of tangent on C1C_1 at (2,1)(2,1) is: 2x+y−(x+2)−1=0Orx+y=32 x+y-(x+2)-1=0 Or x+y=3 If it cuts off the chord of the circle C2C_2 then the equation of the chord is: x+y=3 ∴\therefore \quad distance of the chord from (3,−2)(3,-2) is : d=∣3−2−32∣=2Also,length\left|\frac{3-2-3}{\sqrt{2}}\right|=\sqrt{2} Also, length of the chord isl=4 ∴ radius of  C2=r=(l2)2+d2=(2)2+(2)2=6\begin{aligned} \therefore\text{ radius\text{ of } } C_2=r &=\sqrt{\left(\frac{l}{2}\right)^2+d^2} \\ &=\sqrt{(2)^2+(\sqrt{2})^2}=\sqrt{6} \end{aligned}
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