To find the equation of the image circle of the circle \( x^2 + y^2 - 2x = 0 \) in the line \( x + y - 2 = 0 \), we can follow these steps:
### Step 1: Identify the center and radius of the original circle
The equation of the circle can be rewritten as:
\[
x^2 - 2x + y^2 = 0
\]
This can be compared to the standard form of a circle:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
To find the center, we complete the square for the \( x \) terms:
\[
(x^2 - 2x + 1) + y^2 = 1
\]
This gives us:
\[
(x - 1)^2 + y^2 = 1
\]
From this, we can see that the center of the circle is \( (1, 0) \) and the radius \( r = 1 \).
### Step 2: Find the equation of the line
The equation of the line is given as:
\[
x + y - 2 = 0
\]
This can be rewritten in the form \( ax + by + c = 0 \) where \( a = 1, b = 1, c = -2 \).
### Step 3: Find the image of the center across the line
To find the image of the center \( (1, 0) \) across the line \( x + y - 2 = 0 \), we use the formula for the reflection of a point \( (x_1, y_1) \) across the line \( ax + by + c = 0 \):
\[
h = x_1 - \frac{2a(ax_1 + by_1 + c)}{a^2 + b^2}
\]
\[
k = y_1 - \frac{2b(ax_1 + by_1 + c)}{a^2 + b^2}
\]
Substituting \( (x_1, y_1) = (1, 0) \) and \( (a, b, c) = (1, 1, -2) \):
\[
h = 1 - \frac{2 \cdot 1 \cdot (1 \cdot 1 + 1 \cdot 0 - 2)}{1^2 + 1^2}
\]
\[
= 1 - \frac{2 \cdot 1 \cdot (1 - 2)}{2} = 1 - \frac{2 \cdot (-1)}{2} = 1 + 1 = 2
\]
\[
k = 0 - \frac{2 \cdot 1 \cdot (1 \cdot 1 + 1 \cdot 0 - 2)}{1^2 + 1^2}
\]
\[
= 0 - \frac{2 \cdot 1 \cdot (1 - 2)}{2} = 0 + 1 = 1
\]
Thus, the image of the center is \( (2, 1) \).
### Step 4: Write the equation of the image circle
The image circle has the same radius as the original circle, which is \( r = 1 \). Therefore, the equation of the image circle with center \( (2, 1) \) and radius \( 1 \) is:
\[
(x - 2)^2 + (y - 1)^2 = 1
\]
Expanding this, we get:
\[
(x^2 - 4x + 4) + (y^2 - 2y + 1) = 1
\]
\[
x^2 + y^2 - 4x - 2y + 4 = 1
\]
\[
x^2 + y^2 - 4x - 2y + 3 = 0
\]
### Final Answer
The equation of the image circle is:
\[
x^2 + y^2 - 4x - 2y + 3 = 0
\]
