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AP EAMCET201921 Apr 2019Evening ShiftMathematicsCircleActual

Two circles each of radius 5 units touch each other at ( 1 , 2 ) and 4 x + 3 y = 10 is their common tangent. The equation of that circle among the two given circles, such that some portion of it lies in every quadrant is

Options

  1. Ax 2 + y 2 + 6 x + 2 y + 15 = 0
  2. Bx 2 + y 2 + 2 x + 6 y - 15 = 0
  3. Cx 2 + y 2 + 6 x + 2 y - 15 = 0
  4. Dx 2 + y 2 - 6 x + 2 y - 15 = 0

Correct answer

C. x 2 + y 2 + 6 x + 2 y - 15 = 0

Step-by-step solution

The figure below represents the two circles with the common tangent.  The slope of the common tangent,  m = - 4 3  The slope of the line perpendicular to tangent is,  m ' = tan θ = 3 4  Therefore,  sin θ = 3 5 , cos θ = 4 5  Now,  x = ± 5 × 4 5 + 1 , y = ± 5 × 3 5 + 2 x = ( 5 , - 3 ) ,   y = ( 5 , - 1 )  The coordinates of  C 1 5 , 5  and  C 2 - 3 , - 1  The equations of the required circles is,  ( x -

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