An isosceles right traingle ABC with BC = AC slides on the axes with A and B on x and y axes respectively. The locus of C is …………… .

Answer

To find the locus of point C in the isosceles right triangle ABC, where A is on the x-axis, B is on the y-axis, and C is the right-angle vertex, we can follow these steps:

### Step-by-Step Solution:

1. **Define Points A and B**:
Let the coordinates of point A be \( (h, 0) \) and the coordinates of point B be \( (0, k) \). Here, \( h \) and \( k \) are the lengths of the sides of the triangle along the x-axis and y-axis, respectively.

2. **Identify Point C**:
Since triangle ABC is an isosceles right triangle, the lengths \( AC \) and \( BC \) are equal. Let the coordinates of point C be \( (0, k) \) on the y-axis and \( (h, 0) \) on the x-axis.

3. **Use the Distance Formula**:
The distance \( AC \) can be calculated as:
\[ AC = \sqrt{(h - 0)^2 + (0 - k)^2} = \sqrt{h^2 + k^2} \]
The distance \( BC \) can be calculated as:
\[ BC = \sqrt{(0 - h)^2 + (k - 0)^2} = \sqrt{h^2 + k^2} \]
Since \( AC = BC \), we have:
\[ \sqrt{h^2 + k^2} = \sqrt{h^2 + k^2} \]

4. **Establish the Relationship**:
Since both distances are equal, we can express the relationship between \( h \) and \( k \) using the property of the triangle:
\[ h^2 + k^2 = A^2 \]
where \( A \) is the length of the equal sides of the triangle.

5. **Find the Locus of Point C**:
From the relationship derived, we can express \( k \) in terms of \( h \):
\[ k = \sqrt{A^2 - h^2} \]
This implies that as point C moves, the coordinates \( (h, k) \) will satisfy the equation of a circle with radius \( A \) centered at the origin.

6. **Final Equation**:
The locus of point C can be expressed as:
\[ k^2 + h^2 = A^2 \]
This represents a circle with radius \( A \).

### Conclusion:
The locus of point C is a circle centered at the origin with radius equal to the length of the sides of the triangle.