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AP EAMCET201822 Apr 2018Evening ShiftMathematicsCircleActual

The line x-2=0 cuts the circle x^2+y^2-8 x-2 y+8=0 at A and B . The equation of the circle passing through the points A and B and having least radius is

Options

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

Correct answer

B. x^2+y^2-4 x-2 y=0

Step-by-step solution

Equation of circles passes through the point of intersection of line x-2=0 and the circle x^2+y^2-8 x-2 y+8=0 , is (x^2+y^2-8 x-2 y+8 )+ (x-2)=0 For minimum radius, it is necessary that centre of circle Eq. (i) lies on the line x-2=0 , so gathered - ( -8 2 )-2=0 -8=-4 =4 gathered So, equation of required circle is x^2+y^2-4 x-2 y=0

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