To find the center of the circle passing through the points (0, 0), (2, 0), and (0, 2), we can follow these steps:
### Step 1: Understand the Problem
We need to find the center of a circle that passes through the three given points. The center of the circle will be equidistant from all three points.
### Step 2: Set Up the Distance Equations
Let the center of the circle be at point \( P(x, y) \). We will use the distance formula to set up equations based on the distances from point \( P \) to each of the three points.
1. Distance from \( P \) to \( A(0, 0) \):
\[
PA^2 = (x - 0)^2 + (y - 0)^2 = x^2 + y^2 \quad \text{(Equation 1)}
\]
2. Distance from \( P \) to \( B(2, 0) \):
\[
PB^2 = (x - 2)^2 + (y - 0)^2 = (x - 2)^2 + y^2 = x^2 - 4x + 4 + y^2 \quad \text{(Equation 2)}
\]
3. Distance from \( P \) to \( C(0, 2) \):
\[
PC^2 = (x - 0)^2 + (y - 2)^2 = x^2 + (y - 2)^2 = x^2 + y^2 - 4y + 4 \quad \text{(Equation 3)}
\]
### Step 3: Equate Distances
Since all distances are equal (as they are all radii of the same circle), we can set the equations equal to each other.
#### Equating Equation 1 and Equation 2:
\[
x^2 + y^2 = x^2 - 4x + 4 + y^2
\]
Cancelling \( x^2 \) and \( y^2 \) from both sides gives:
\[
0 = -4x + 4 \implies 4x = 4 \implies x = 1 \quad \text{(Equation 4)}
\]
#### Equating Equation 2 and Equation 3:
\[
x^2 - 4x + 4 + y^2 = x^2 + y^2 - 4y + 4
\]
Cancelling \( x^2 \) and \( y^2 \) from both sides gives:
\[
-4x + 4 = -4y + 4 \implies -4x = -4y \implies x = y \quad \text{(Equation 5)}
\]
### Step 4: Solve for \( y \)
From Equation 4, we have \( x = 1 \). Substituting \( x \) into Equation 5:
\[
1 = y \implies y = 1
\]
### Step 5: Conclusion
Thus, the center of the circle is at the point \( (1, 1) \).
### Final Answer
The center of the circle passing through the points \( (0, 0) \), \( (2, 0) \), and \( (0, 2) \) is \( (1, 1) \).
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