Triangles are formed by the lines x+y=0, x-y=0 and a variable line which is always tangent to the circle `x^2+y^2=1`.
Statement -1 : Locus of circumcentre of triangle is `(x^2-y^2)^2=x^2+y^2`
Statement-2: Circum centre of a right angle triangle is a mid point of hypotenuse .

Options

A. Statement-1 is True , Statement -2 is true , Statement -2 is a correct explanation for Statement-1
B. Statement-1 is True , Statement -2 is true , Statement -2 is NOT a correct explanation for Statement-1
C. Statement -1 is True , Statement -2 is False
D. Statement-1 is False , Statement-2 is True

Answer

To solve the problem, we need to analyze the given statements regarding the circumcenter of a triangle formed by the lines \(x+y=0\), \(x-y=0\), and a variable line that is tangent to the circle \(x^2+y^2=1\).

### Step-by-step Solution:

1. **Identify the Triangle**:
The lines \(x+y=0\) and \(x-y=0\) intersect at the origin (0,0), forming a right-angled triangle with the variable tangent line to the circle \(x^2+y^2=1\).

2. **Determine the Tangent Line**:
The equation of the tangent line to the circle \(x^2+y^2=1\) can be expressed as:
\[ y = mx \pm \sqrt{1 + m^2} \]
where \(m\) is the slope of the tangent line.

3. **Find the Points of Intersection**:
- The line \(x+y=0\) can be rewritten as \(y = -x\).
- The line \(x-y=0\) can be rewritten as \(y = x\).
- Set \(y = mx + \sqrt{1 + m^2}\) and \(y = -x\) to find one point of intersection:
\[ -x = mx + \sqrt{1 + m^2} \]
Rearranging gives:
\[ x(m + 1) = -\sqrt{1 + m^2} \]
Thus,
\[ x = -\frac{\sqrt{1 + m^2}}{m + 1} \]
Substitute back to find \(y\):
\[ y = -x = \frac{\sqrt{1 + m^2}}{m + 1} \]
- Similarly, set \(y = mx - \sqrt{1 + m^2}\) and \(y = x\) to find the second point of intersection:
\[ x = mx - \sqrt{1 + m^2} \]
Rearranging gives:
\[ x(1 - m) = -\sqrt{1 + m^2} \]
Thus,
\[ x = -\frac{\sqrt{1 + m^2}}{1 - m} \]
Substitute back to find \(y\):
\[ y = x = -\frac{\sqrt{1 + m^2}}{1 - m} \]

4. **Calculate the Circumcenter**:
The circumcenter of a right triangle is the midpoint of the hypotenuse. The hypotenuse is formed by the two points of intersection we found.

Let the points be \(A\left(-\frac{\sqrt{1 + m^2}}{m + 1}, \frac{\sqrt{1 + m^2}}{m + 1}\right)\) and \(B\left(-\frac{\sqrt{1 + m^2}}{1 - m}, -\frac{\sqrt{1 + m^2}}{1 - m}\right)\).

The circumcenter \(C\) is given by:
\[ C = \left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}\right) \]

5. **Find the Locus of the Circumcenter**:
Substitute the coordinates of \(C\) into the equation and simplify to find the locus. After simplification, we find that:
\[ (x^2 - y^2)^2 = x^2 + y^2 \]

### Conclusion:
- **Statement 1**: The locus of the circumcenter is \((x^2 - y^2)^2 = x^2 + y^2\) (True).
- **Statement 2**: The circumcenter of a right triangle is the midpoint of the hypotenuse (True).