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JEE Main 20207 JanuaryEvening Shift
Exam
JEE Main
Year
2020
Date
7 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Circle

JEE Main 2020 · 7 January · Evening Shift · Mathematics · Circle

Let the tangents drawn from the origin to the circle, x ^ 2 + y ^ 2...

Let the tangents drawn from the origin to the circle, x2+y2−8x−4y+16=0{x}^{2} + {y}^{2} - 8x - 4y + 16 = 0 touch it at the points AandA andB.Then. Then{\left(AB\right)}^{2} is equal to

Options

  1. A525\dfrac{52}{5}
  2. B565\dfrac{56}{5}
  3. C645\dfrac{64}{5}
  4. D325\dfrac{32}{5}

Correct answer

C. 645\dfrac{64}{5}

Step-by-step solution

L=S1=16=4L = \sqrt{{S}_{1}} = \sqrt{16} = 4
R=16+4−16=2R = \sqrt{16 + 4 - 16} = 2
Length of chord of contact =2LRL2+R2=2×4×216+4=1620= \dfrac{2LR}{\sqrt{{L}^{2} + {R}^{2}}} = \dfrac{2 \times 4 \times 2}{\sqrt{16 + 4}} = \dfrac{16}{\sqrt{20}}
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