A triangle is formed by the lines `x+y=0,x-y=0,` and `l x+m y=1.` If `la n dm` vary subject to the condition `l^2+m^2=1,` then the locus of its circumcenter is (a) `(x^2-y^2)^2=x^2+y^2` (b) `(x^2+y^2)^2=(x^2-y^2)` (c) `(x^2+y^2)^2=4x^2y^2` (d) `(x^2-y^2)^2=(x^2+y^2)^2`

Options

A. `(x^(2)-y^(2))^(2) = x^(2) + y^(2)`
B. `(x^(2)-y^(2))^(2) = (x^(2) - y^(2))`
C. `(x^(2)-y^(2)) = 4x^(2)y^(2)`
D. `(x^(2)-y^(2))^(2) = (x^(2)+y^(2))^(2)`

Explanation

To find the locus of the circumcenter of the triangle formed by the lines \( x + y = 0 \), \( x - y = 0 \), and \( l x + m y = 1 \) under the condition \( l^2 + m^2 = 1 \), we can follow these steps:

### Step 1: Identify the vertices of the triangle
The lines \( x + y = 0 \) and \( x - y = 0 \) intersect at the origin (0, 0).

To find the intersection points of the line \( l x + m y = 1 \) with the other two lines:
1. Substitute \( y = -x \) into \( l x + m(-x) = 1 \):
\[ (l - m)x = 1 \implies x = \frac{1}{l - m}, \quad y = -\frac{1}{l - m} \]
So, one vertex is \( \left(\frac{1}{l - m}, -\frac{1}{l - m}\right) \).

2. Substitute \( y = x \) into \( l x + m x = 1 \):
\[ (l + m)x = 1 \implies x = \frac{1}{l + m}, \quad y = \frac{1}{l + m} \]
So, another vertex is \( \left(\frac{1}{l + m}, \frac{1}{l + m}\right) \).

Thus, the vertices of the triangle are:
- \( A(0, 0) \)
- \( B\left(\frac{1}{l - m}, -\frac{1}{l - m}\right) \)
- \( C\left(\frac{1}{l + m}, \frac{1}{l + m}\right) \)

### Step 2: Find the circumcenter
The circumcenter of a right triangle is located at the midpoint of the hypotenuse. The hypotenuse is the segment connecting points \( B \) and \( C \).

The midpoint \( O \) of \( BC \) is given by:
\[ O = \left( \frac{\frac{1}{l - m} + \frac{1}{l + m}}{2}, \frac{-\frac{1}{l - m} + \frac{1}{l + m}}{2} \right) \]

### Step 3: Simplify the coordinates of the circumcenter
Calculating the x-coordinate:
\[ x = \frac{1}{2} \left( \frac{1}{l - m} + \frac{1}{l + m} \right) = \frac{(l + m) + (l - m)}{2(l^2 - m^2)} = \frac{2l}{2(l^2 - m^2)} = \frac{l}{l^2 - m^2} \]

Calculating the y-coordinate:
\[ y = \frac{1}{2} \left( -\frac{1}{l - m} + \frac{1}{l + m} \right) = \frac{-(l + m) + (l - m)}{2(l^2 - m^2)} = \frac{-2m}{2(l^2 - m^2)} = \frac{-m}{l^2 - m^2} \]

Thus, the circumcenter \( O \) has coordinates:
\[ O\left(\frac{l}{l^2 - m^2}, \frac{-m}{l^2 - m^2}\right) \]

### Step 4: Find the locus
To find the locus, we express \( x \) and \( y \) in terms of \( l \) and \( m \):
\[ x = \frac{l}{l^2 - m^2}, \quad y = \frac{-m}{l^2 - m^2} \]

From \( l^2 + m^2 = 1 \), we can express \( l \) and \( m \) in terms of \( x \) and \( y \) and substitute back to eliminate \( l \) and \( m \).

After substituting and simplifying, we arrive at the equation:
\[ (x^2 - y^2)^2 = x^2 + y^2 \]

### Conclusion
The locus of the circumcenter of the triangle is given by:
\[ \boxed{(x^2 - y^2)^2 = x^2 + y^2} \]