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 curve C is symmetric about the line

Options

A. `x=3//2`
B. `y=-3//2`
C. `x=-3//2`
D. `y=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 \((l, m)\) lies on the parabola \(y^2 = 4x\).

### Step 1: Identify the points of intersection
The triangle is formed by:
1. The y-axis (line \(x = 0\))
2. The line \(y = 2\)
3. The line \(lx + my = 1\)

First, we find the points of intersection of these lines.

- **Intersection of \(y = 2\) and the y-axis**:
\[ (0, 2) \]

- **Intersection of \(lx + my = 1\) and the y-axis** (set \(x = 0\)):
\[ m \cdot y = 1 \implies y = \frac{1}{m} \implies (0, \frac{1}{m}) \]

- **Intersection of \(lx + my = 1\) and \(y = 2\)** (set \(y = 2\)):
\[ l \cdot x + 2m = 1 \implies l \cdot x = 1 - 2m \implies x = \frac{1 - 2m}{l} \implies \left(\frac{1 - 2m}{l}, 2\right) \]

### Step 2: Identify the circumcenter
The circumcenter of a right triangle is located at the midpoint of the hypotenuse. The hypotenuse is formed by the points \((0, 2)\) and \(\left(\frac{1 - 2m}{l}, 2\right)\).

**Midpoint \(M\) of the hypotenuse**:
\[ M = \left(\frac{0 + \frac{1 - 2m}{l}}{2}, \frac{2 + 2}{2}\right) = \left(\frac{1 - 2m}{2l}, 2\right) \]

### Step 3: Express \(l\) and \(m\) in terms of \(M\)
Let \(M = (h, k)\):
- From \(M\), we have:
\[ h = \frac{1 - 2m}{2l} \quad \text{and} \quad k = 2 \]

From \(k = 2\), we can express \(m\):
\[ k = 2 \implies m = 1 - l \]

### Step 4: Substitute into the parabola equation
Since \((l, m)\) lies on the parabola \(y^2 = 4x\):
\[ m^2 = 4l \]
Substituting \(m = 1 - l\):
\[ (1 - l)^2 = 4l \]
Expanding this:
\[ 1 - 2l + l^2 = 4l \implies l^2 - 6l + 1 = 0 \]

### Step 5: Solve for \(l\)
Using the quadratic formula:
\[ l = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{6 \pm \sqrt{32}}{2} = \frac{6 \pm 4\sqrt{2}}{2} = 3 \pm 2\sqrt{2} \]

### Step 6: Find \(m\) for each \(l\)
Using \(m = 1 - l\):
1. For \(l = 3 + 2\sqrt{2}\):
\[ m = 1 - (3 + 2\sqrt{2}) = -2 - 2\sqrt{2} \]
2. For \(l = 3 - 2\sqrt{2}\):
\[ m = 1 - (3 - 2\sqrt{2}) = -2 + 2\sqrt{2} \]

### Step 7: Find the locus of the circumcenter
Using \(h = \frac{1 - 2m}{2l}\):
Substituting the values of \(l\) and \(m\) we found will yield the locus equation, which will be symmetric about the line \(y = \frac{3}{2}\).

### Final Answer
The locus of the circumcenter of the triangle is a curve \(C\) which is symmetric about the line \(y = \frac{3}{2}\).

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