If the line y = mx + 1 meets the circle x^2 + | vedclass

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(B) The equation of the circle is $x^2 + y^2 + 3x = 0$.Substituting $y = mx + 1$ into the circle equation:$x^2 + (mx + 1)^2 + 3x = 0$$x^2 + m^2x^2 + 2

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$2m + 3 = 0$

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(B) The equation of the circle is $x^2 + y^2 + 3x = 0$.Substituting $y = mx + 1$ into the circle equation:$x^2 + (mx + 1)^2 + 3x = 0$$x^2 + m^2x^2 + 2mx + 1 + 3x = 0$$(1 + m^2)x^2 + (2m + 3)x + 1 = 0$.Let the roots of this quadratic equation be $x_1$ and $x_2$. These represent the $x$-coordinates of the intersection points.Since the points are equidistant from the $y$-axis and on opposite sides,we must have $x_1 = -x_2$,which implies $x_1 + x_2 = 0$.For a quadratic equation $ax^2 + bx + c = 0$,the sum of roots is $-b/a$.Thus,$-(2m + 3) / (1 + m^2) = 0$.This implies $2m + 3 = 0$.

If the line $y = mx + 1$ meets the circle $x^2 + y^2 + 3x = 0$ at two points equidistant from the $y$-axis and on opposite sides,then:

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