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MHT CET202120 Sep 2021Evening ShiftMathematicsCircleActual

The equation of tangent to the circle x^2+y^2=64 at the point P ( 2 3 ) is

Options

  1. Ax- 3 y-16=0
  2. B3 x + y -16=0
  3. Cx + 3 y +16=0
  4. Dx - 3 y +16=0

Correct answer

D. x - 3 y +16=0

Step-by-step solution

Circle x^2+y^2=(8)^2 , has radius 8 and centre (0,0) . Point P ( 2 3 ) on the circle has coordinates P (8 2 3 , 8 2 3 ) i.e. P (-4,4 3 ) Differentiating equation of circle w.r.t. x , we get 2 x+2 y d y d x =0 d y d x = -x y ( d y d x )_P= 4 4 3 = 1 3 Hence required equation of tangent is ( y -4 3 )= 1 3 ( x +4) x - 3 y +16=0

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