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AP EAMCET202316 May 2023Evening ShiftMathematicsCircleActual

The equation of a tangent to the circle x^2+y^2+2 x-12 y-132=0 which is perpendicular to the line 12 x+5 y+k=0 is

Options

  1. A5 x-12 y+92=0
  2. B5 x -12 y -246=0
  3. C5 x -12 y -169=0
  4. D5 x-12 y+246=0

Correct answer

D. 5 x-12 y+246=0

Step-by-step solution

Given equation of circle aligned & x^2+y^2+2 x-12 y-132=0 & (x+1)^2+(y-6)^2=13^2 & Radius =13, Centre =(-1,6) aligned Since slope of the line 12 x+5 y+8=0 is m ₁= -12 5 So slope of perpendicular line to the given line is m ₂= 5 12 Now, equation of required line is y= 5 12 x+C Since, perpendicular distance of tangent from centre =13 aligned & | 5 12 (-1)+C-6 | ( 5 12 )^2+1 =13 & |-77+12 C|=169 C=- 92 12 or 246 12 aligned so, y= 5 x 12 - 92 12 5 x-12 y-92=0 or, y= 5 12 x+ 246 12 5 x-12 y+246=0

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