MathematicsMedium127×since 2002Q3027The points of intersection of the line ax+by=0,(a≠b)ax + by = 0,(a \ne b)ax+by=0,(a=b) and the circle x2+y2−2x=0{x^2} + {y^2} - 2x = 0x2+y2−2x=0 are A(α,0)A(\alpha ,0)A(α,0) and B(1,β)B(1,\beta )B(1,β). The image of the circle with AB as a diameter in the line x+y+2=0x + y + 2 = 0x+y+2=0 is :Ax2+y2+5x+5y+12=0{x^2} + {y^2} + 5x + 5y + 12 = 0x2+y2+5x+5y+12=0Bx2+y2+3x+5y+8=0{x^2} + {y^2} + 3x + 5y + 8 = 0x2+y2+3x+5y+8=0Cx2+y2−5x−5y+12=0{x^2} + {y^2} - 5x - 5y + 12 = 0x2+y2−5x−5y+12=0Dx2+y2+3x+3y+4=0{x^2} + {y^2} + 3x + 3y + 4 = 0x2+y2+3x+3y+4=0Check answerSkip