The minimum value of k for which `|z-i|^2 + |z-1|^2 =k` will represent a circle is ____________.

Answer

To find the minimum value of \( k \) for which the equation \( |z - i|^2 + |z - 1|^2 = k \) represents a circle, we will follow these steps:

1. **Substituting \( z \)**: We start by letting \( z = x + iy \), where \( x \) and \( y \) are real numbers.

2. **Expanding the terms**: We need to expand \( |z - i|^2 \) and \( |z - 1|^2 \):
- \( |z - i|^2 = |(x + iy) - i|^2 = |x + i(y - 1)|^2 = x^2 + (y - 1)^2 \)
- \( |z - 1|^2 = |(x + iy) - 1|^2 = |(x - 1) + iy|^2 = (x - 1)^2 + y^2 \)

3. **Combining the equations**: Now we combine both expressions:
\[ |z - i|^2 + |z - 1|^2 = (x^2 + (y - 1)^2) + ((x - 1)^2 + y^2) \]
Expanding this gives:
\[ = x^2 + (y^2 - 2y + 1) + (x^2 - 2x + 1 + y^2) \]
\[ = 2x^2 + 2y^2 - 2x - 2y + 2 \]

4. **Setting the equation**: We set this equal to \( k \):
\[ 2x^2 + 2y^2 - 2x - 2y + 2 = k \]

5. **Rearranging the equation**: Rearranging gives:
\[ 2x^2 + 2y^2 - 2x - 2y + (2 - k) = 0 \]

6. **Dividing by 2**: To simplify, we divide everything by 2:
\[ x^2 + y^2 - x - y + \frac{2 - k}{2} = 0 \]

7. **Completing the square**: We complete the square for \( x \) and \( y \):
\[ (x^2 - x) + (y^2 - y) = \frac{k - 2}{2} \]
Completing the square:
\[ \left(x - \frac{1}{2}\right)^2 - \frac{1}{4} + \left(y - \frac{1}{2}\right)^2 - \frac{1}{4} = \frac{k - 2}{2} \]
This simplifies to:
\[ \left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 = \frac{k - 2}{2} + \frac{1}{2} \]
\[ = \frac{k - 1}{2} \]

8. **Condition for a circle**: For this to represent a circle, the right-hand side must be non-negative:
\[ \frac{k - 1}{2} \geq 0 \]
This implies:
\[ k - 1 \geq 0 \quad \Rightarrow \quad k \geq 1 \]

9. **Conclusion**: The minimum value of \( k \) for which the equation represents a circle is:
\[ \boxed{1} \]