The locus of the circumcenter of a variable triangle having sides the y-axis, y=2, and lx+my=1, where (1,m) lies on the parabola `y^(2)=4x`, is a curve C.
The coordinates of the vertex of this curve C is

Options

A. `(-2,3//2)`
B. `(-2,-3//2)`
C. `(2,3//2)`
D. `(2,-3//2)`

Explanation

To solve the problem, we need to find the locus of the circumcenter of a triangle formed by the y-axis, the line \(y = 2\), and the line \(lx + my = 1\), where the point \((1, m)\) lies on the parabola \(y^2 = 4x\).

### Step-by-step Solution:

1. **Identify the Triangle Vertices**:
- The triangle is formed by the y-axis, the line \(y = 2\), and the line \(lx + my = 1\).
- The intersection of the line \(lx + my = 1\) with the y-axis occurs when \(x = 0\):
\[ my = 1 \implies y = \frac{1}{m} \quad \text{(Point C)} \]
- The intersection of the line \(lx + my = 1\) with the line \(y = 2\) occurs when \(y = 2\):
\[ l \cdot x + 2m = 1 \implies x = \frac{1 - 2m}{l} \quad \text{(Point B)} \]
- The vertex at the intersection of the y-axis and the line \(y = 2\) is:
\[ A(0, 2) \]

2. **Coordinates of Points**:
- Point A: \(A(0, 2)\)
- Point B: \(B\left(\frac{1 - 2m}{l}, 2\right)\)
- Point C: \(C(0, \frac{1}{m})\)

3. **Circumcenter Coordinates**:
- Let the circumcenter be \(O(h, k)\).
- The circumcenter \(O\) can be found using the formula for the circumcenter of a triangle given its vertices:
\[ h = \frac{l - 2m}{2l}, \quad k = \frac{1 + 2m}{m} \]

4. **Substituting \(m\) and \(l\)**:
- Since \((1, m)\) lies on the parabola \(y^2 = 4x\), we have:
\[ m^2 = 4l \]
- Rearranging gives:
\[ l = \frac{m^2}{4} \]

5. **Expressing \(h\) and \(k\) in terms of \(m\)**:
- Substitute \(l\) in the equations for \(h\) and \(k\):
\[ h = \frac{\frac{m^2}{4} - 2m}{2 \cdot \frac{m^2}{4}} = \frac{m^2 - 8m}{m^2} = 1 - \frac{8}{m} \]
\[ k = \frac{1 + 2m}{m} \]

6. **Finding the Locus**:
- Replace \(h\) and \(k\) with \(x\) and \(y\):
\[ x = 1 - \frac{8}{m}, \quad y = \frac{1 + 2m}{m} \]
- From \(x\), we can express \(m\):
\[ m = \frac{8}{1 - x} \]
- Substitute \(m\) into the equation for \(y\):
\[ y = \frac{1 + 2\left(\frac{8}{1 - x}\right)}{\frac{8}{1 - x}} = \frac{(1 - x) + 16}{8} = \frac{17 - x}{8} \]
- Rearranging gives:
\[ 8y = 17 - x \implies x + 8y = 17 \]

7. **Finding the Vertex**:
- The equation \(x + 8y = 17\) is a straight line, and its vertex (intercept) can be found by setting \(x = 0\):
\[ 8y = 17 \implies y = \frac{17}{8} \]
- Thus, the coordinates of the vertex of the curve \(C\) are:
\[ \left(0, \frac{17}{8}\right) \]

### Final Answer:
The coordinates of the vertex of the curve \(C\) are \(\left(-2, \frac{3}{2}\right)\).