If the line y=mx +1 meets the circle `x^(2)+y^(2)+3x=0` in two points equidistant and on opposite sides of x-axis, then

Options

A. 3m-2=0
B. 2m+3=0
C. 3m+2=0
D. 2m-3=0

Explanation

To solve the problem, we need to find the value of \( m \) such that the line \( y = mx + 1 \) intersects the circle defined by the equation \( x^2 + y^2 + 3x = 0 \) at two points that are equidistant from and on opposite sides of the x-axis.

### Step-by-Step Solution:

1. **Rewrite the Circle Equation**:
The given circle equation is:
\[ x^2 + y^2 + 3x = 0 \]
We can rearrange this into standard form:
\[ x^2 + 3x + y^2 = 0 \]
Completing the square for the \( x \) terms:
\[ (x + \frac{3}{2})^2 - \frac{9}{4} + y^2 = 0 \]
This simplifies to:
\[ (x + \frac{3}{2})^2 + y^2 = \frac{9}{4} \]
This represents a circle centered at \( (-\frac{3}{2}, 0) \) with a radius of \( \frac{3}{2} \).

2. **Substitute the Line Equation into the Circle**:
Substitute \( y = mx + 1 \) into the circle's equation:
\[ x^2 + (mx + 1)^2 + 3x = 0 \]
Expanding \( (mx + 1)^2 \):
\[ x^2 + (m^2x^2 + 2mx + 1) + 3x = 0 \]
Combine like terms:
\[ (1 + m^2)x^2 + (2m + 3)x + 1 = 0 \]

3. **Condition for Roots**:
For the line to intersect the circle at two points that are equidistant from and on opposite sides of the x-axis, the sum of the roots of the quadratic equation must be zero. The sum of the roots \( \alpha \) and \( -\alpha \) gives:
\[ \text{Sum of roots} = -\frac{b}{a} = 0 \]
Here, \( a = 1 + m^2 \) and \( b = 2m + 3 \). Therefore:
\[ -\frac{2m + 3}{1 + m^2} = 0 \]
This implies:
\[ 2m + 3 = 0 \]

4. **Solve for \( m \)**:
Solving the equation \( 2m + 3 = 0 \):
\[ 2m = -3 \quad \Rightarrow \quad m = -\frac{3}{2} \]

5. **Conclusion**:
The value of \( m \) for which the line intersects the circle at two points that are equidistant and on opposite sides of the x-axis is:
\[ m = -\frac{3}{2} \]