To solve the problem, we need to find the distance between the centers of two intersecting circles that share the same radius and intersect at the points (0, 1) and (0, -1).
### Step-by-Step Solution:
1. **Define the Circles**:
Let the centers of the two circles be \( (a, b) \) and \( (c, d) \) with equal radius \( r \). The equations of the circles can be written as:
\[
(x - a)^2 + (y - b)^2 = r^2 \quad \text{(Circle 1)}
\]
\[
(x - c)^2 + (y - d)^2 = r^2 \quad \text{(Circle 2)}
\]
2. **Substituting Intersection Points**:
Since both circles intersect at the points (0, 1) and (0, -1), we can substitute these points into the equations of both circles.
For Circle 1 at (0, 1):
\[
(0 - a)^2 + (1 - b)^2 = r^2 \quad \text{(1)}
\]
For Circle 1 at (0, -1):
\[
(0 - a)^2 + (-1 - b)^2 = r^2 \quad \text{(2)}
\]
For Circle 2 at (0, 1):
\[
(0 - c)^2 + (1 - d)^2 = r^2 \quad \text{(3)}
\]
For Circle 2 at (0, -1):
\[
(0 - c)^2 + (-1 - d)^2 = r^2 \quad \text{(4)}
\]
3. **Equating the Radius Expressions**:
From equations (1) and (2):
\[
a^2 + (1 - b)^2 = a^2 + (-1 - b)^2
\]
Simplifying gives:
\[
(1 - b)^2 = (-1 - b)^2
\]
Expanding both sides:
\[
1 - 2b + b^2 = 1 + 2b + b^2
\]
Canceling \( b^2 \) and 1 from both sides:
\[
-2b = 2b \implies 4b = 0 \implies b = 0
\]
4. **Finding the Center of Circle 1**:
Now substituting \( b = 0 \) back into the equation for Circle 1:
\[
(0 - a)^2 + (1 - 0)^2 = r^2 \implies a^2 + 1 = r^2 \quad \text{(5)}
\]
For Circle 1 at (0, -1):
\[
(0 - a)^2 + (-1 - 0)^2 = r^2 \implies a^2 + 1 = r^2 \quad \text{(same as 5)}
\]
5. **Finding the Center of Circle 2**:
Similarly, substituting \( d = 0 \) into Circle 2:
\[
(0 - c)^2 + (1 - 0)^2 = r^2 \implies c^2 + 1 = r^2 \quad \text{(6)}
\]
For Circle 2 at (0, -1):
\[
(0 - c)^2 + (-1 - 0)^2 = r^2 \implies c^2 + 1 = r^2 \quad \text{(same as 6)}
\]
6. **Relating the Centers**:
Since the tangent at (0, 1) of Circle 1 passes through the center of Circle 2, we can find the relationship between \( a \) and \( c \). The tangent slope at (0, 1) is given by:
\[
\text{slope} = \frac{a - 0}{0 - 1} = -a
\]
The equation of the tangent line at (0, 1) is:
\[
y - 1 = -a(x - 0) \implies y = -ax + 1
\]
This line passes through the center of Circle 2, \( (c, 0) \):
\[
0 = -ac + 1 \implies ac = 1 \quad \text{(7)}
\]
7. **Finding the Distance Between Centers**:
From equations (5) and (6), we have:
\[
r^2 = a^2 + 1 = c^2 + 1 \implies a^2 = c^2
\]
This gives us two cases:
\[
a = c \quad \text{or} \quad a = -c
\]
Since the circles are distinct, we take \( a = -c \).
The distance \( d \) between the centers \( (a, 0) \) and \( (c, 0) \) is:
\[
d = |a - c| = |a - (-a)| = |2a|
\]
8. **Finding the Value of \( a \)**:
From \( ac = 1 \) and \( a = -c \):
\[
a(-a) = 1 \implies -a^2 = 1 \implies a^2 = 1 \implies a = 1 \text{ or } -1
\]
9. **Final Distance Calculation**:
Thus, the distance between the centers is:
\[
d = |2a| = |2 \cdot 1| = 2 \quad \text{or} \quad |2 \cdot (-1)| = 2
\]
### Final Answer:
The distance between the centers of the two circles is \( \boxed{2} \).
