MathematicsEasy127×since 2002Q3018The point diametrically opposite to the point P(1,0)P(1, 0)P(1,0) on the circle x2+y2+2x+4y−3=0{x^2} + {y^2} + 2x + 4y - 3 = 0x2+y2+2x+4y−3=0 is :A(3,−4)(3, -4)(3,−4)B(−3,4)(-3, 4)(−3,4)C(−3,−4)(-3, -4)(−3,−4)D(3,4)(3, 4)(3,4)Check answerSkip