MathematicsMedium92×since 2002Q4703Let PPP be the point on the parabola, y2=8x{{y^2} = 8x}y2=8x which is at a minimum distance from the centre CCC of the circle, x2+(y+6)2=1{x^2} + {\left( {y + 6} \right)^2} = 1x2+(y+6)2=1. Then the equation of the circle, passing through CCC and having its centre at PPP is:Ax2+y2−x4+2y−24=0{{x^2} + {y^2} - {x \over 4} + 2y - 24 = 0}x2+y2−4x+2y−24=0Bx2+y2−4x+9y+18=0{{x^2} + {y^2} - 4x + 9y + 18 = 0}x2+y2−4x+9y+18=0Cx2+y2−4x+8y+12=0{{x^2} + {y^2} - 4x + 8y + 12 = 0}x2+y2−4x+8y+12=0Dx2+y2−x+4y−12=0{{x^2} + {y^2} - x + 4y - 12 = 0}x2+y2−x+4y−12=0Check answerSkip