Find the equations to the circles which pass through the points
`(a, b) , (a, - b) , and (a + b, a - b)`

Answer

To find the equations of the circles that pass through the points \( (a, b) \), \( (a, -b) \), and \( (a + b, a - b) \), we can follow these steps:

### Step 1: Write the general equation of a circle
The general equation of a circle can be expressed as:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
where \( g \), \( f \), and \( c \) are constants that we need to determine.

### Step 2: Substitute the first point \( (a, b) \)
Substituting the point \( (a, b) \) into the circle equation gives:
\[ a^2 + b^2 + 2ga + 2fb + c = 0 \tag{1} \]

### Step 3: Substitute the second point \( (a, -b) \)
Substituting the point \( (a, -b) \) into the circle equation gives:
\[ a^2 + (-b)^2 + 2ga + 2f(-b) + c = 0 \]
This simplifies to:
\[ a^2 + b^2 + 2ga - 2fb + c = 0 \tag{2} \]

### Step 4: Subtract equations (1) and (2)
Subtracting equation (1) from equation (2):
\[ (a^2 + b^2 + 2ga - 2fb + c) - (a^2 + b^2 + 2ga + 2fb + c) = 0 \]
This simplifies to:
\[ -4fb = 0 \]
Thus, we conclude that:
\[ f = 0 \]

### Step 5: Update the circle equation
With \( f = 0 \), the circle equation simplifies to:
\[ x^2 + y^2 + 2gx + c = 0 \]

### Step 6: Substitute the third point \( (a + b, a - b) \)
Now, substituting the point \( (a + b, a - b) \):
\[ (a + b)^2 + (a - b)^2 + 2g(a + b) + c = 0 \]
Expanding this:
\[ (a^2 + 2ab + b^2) + (a^2 - 2ab + b^2) + 2g(a + b) + c = 0 \]
This simplifies to:
\[ 2a^2 + 2b^2 + 2g(a + b) + c = 0 \]
Dividing through by 2:
\[ a^2 + b^2 + g(a + b) + \frac{c}{2} = 0 \tag{3} \]

### Step 7: Use equation (1) to find \( c \)
From equation (1):
\[ a^2 + b^2 + 2ga + c = 0 \]
Rearranging gives:
\[ c = - (a^2 + b^2 + 2ga) \tag{4} \]

### Step 8: Substitute \( c \) into equation (3)
Substituting equation (4) into equation (3):
\[ a^2 + b^2 + g(a + b) - \frac{(a^2 + b^2 + 2ga)}{2} = 0 \]
Multiplying through by 2 to eliminate the fraction:
\[ 2(a^2 + b^2) + 2g(a + b) - (a^2 + b^2 + 2ga) = 0 \]
This simplifies to:
\[ a^2 + b^2 + 2g(b) = 0 \]
Rearranging gives:
\[ 2g(a + b) = - (a^2 + b^2) \]
Thus:
\[ g = -\frac{a^2 + b^2}{2(a + b)} \tag{5} \]

### Step 9: Substitute \( g \) and \( c \) back into the circle equation
Now substituting \( g \) from equation (5) and \( c \) from equation (4) back into the circle equation:
\[ x^2 + y^2 - \frac{a^2 + b^2}{a + b}(x + y) - (a^2 + b^2 + 2g a) = 0 \]
This gives us the final equation of the circle.

### Final Circle Equation
The final equation of the circle can be expressed as:
\[ b(x^2 + y^2) - (a^2 + b^2)x + (a^2 + b^2)(a - b) = 0 \]