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AP EAMCET201921 Apr 2019Morning ShiftMathematicsCircleActual

The equation of the circle whose diameter is the common chord of the circles x^2+y^2+2 x+3 y+1=0 and x^2+y^2+4 x+3 y+2=0 is

Options

  1. A2 x^2+2 y^2+x+3 y+2=0
  2. B2 x^2+2 y^2+2 x+6 y+1=0
  3. C2 x^2+2 y^2+4 x-3 y-1=0
  4. Dx^2+y^2+2 x+6 y-2=0

Correct answer

B. 2 x^2+2 y^2+2 x+6 y+1=0

Step-by-step solution

The equation of the common chord of the circles x^2+y^2+2 x+3 y+1=0 and x^2+y^2+4 x+3 y+2=0 is given by 2 x+1=0 [using: S₁-S₂=0 ] The equation of a circle passing through the intersection of the given circles is gathered (x^2+y^2+2 x+3 y+1 ) + (x^2+y^2+4 x+3 y+2 )=0 x^2(1+ )+y^2(1+ )+(1+2 ) 2 x+3 y(1+ )+1+2 =0 gathered Since, 2 x+1=0 is a diameter of this circle. Therefore, its centre (- 2 +1 +1 ,- 3 2 ) lies on it array ll & -2 ( 2 +1 +1 )+1=0 & -4 -2+ +1=0 -3 -1=0 & =- 1 3 array On putting =- 1 3 in Eq. (i), we g

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