To find the equation of the circle passing through the points (5, -5), (1, 7), and (-7, 1), we can use the general equation of a circle:
\[
x^2 + y^2 + 2gx + 2fy + c = 0
\]
### Step 1: Substitute the first point (5, -5)
Substituting \( (x, y) = (5, -5) \):
\[
5^2 + (-5)^2 + 2g(5) + 2f(-5) + c = 0
\]
Calculating:
\[
25 + 25 + 10g - 10f + c = 0
\]
This simplifies to:
\[
10g - 10f + c + 50 = 0 \quad \text{(Equation 1)}
\]
### Step 2: Substitute the second point (1, 7)
Substituting \( (x, y) = (1, 7) \):
\[
1^2 + 7^2 + 2g(1) + 2f(7) + c = 0
\]
Calculating:
\[
1 + 49 + 2g + 14f + c = 0
\]
This simplifies to:
\[
2g + 14f + c + 50 = 0 \quad \text{(Equation 2)}
\]
### Step 3: Substitute the third point (-7, 1)
Substituting \( (x, y) = (-7, 1) \):
\[
(-7)^2 + 1^2 + 2g(-7) + 2f(1) + c = 0
\]
Calculating:
\[
49 + 1 - 14g + 2f + c = 0
\]
This simplifies to:
\[
-14g + 2f + c + 50 = 0 \quad \text{(Equation 3)}
\]
### Step 4: Solve the system of equations
Now we have three equations:
1. \( 10g - 10f + c + 50 = 0 \) (Equation 1)
2. \( 2g + 14f + c + 50 = 0 \) (Equation 2)
3. \( -14g + 2f + c + 50 = 0 \) (Equation 3)
We will first eliminate \( c \) by subtracting Equation 3 from Equation 2:
\[
(2g + 14f + c + 50) - (-14g + 2f + c + 50) = 0
\]
This simplifies to:
\[
2g + 14f + c + 50 + 14g - 2f - c - 50 = 0
\]
Combining like terms:
\[
16g + 12f = 0
\]
This gives us:
\[
4g + 3f = 0 \quad \text{(Equation 4)}
\]
### Step 5: Substitute \( g \) in terms of \( f \)
From Equation 4, we can express \( g \):
\[
g = -\frac{3}{4}f
\]
### Step 6: Substitute \( g \) in Equation 1
Now substitute \( g \) into Equation 1:
\[
10\left(-\frac{3}{4}f\right) - 10f + c + 50 = 0
\]
This simplifies to:
\[
-\frac{30}{4}f - 10f + c + 50 = 0
\]
Multiplying through by 4 to eliminate the fraction:
\[
-30f - 40f + 4c + 200 = 0
\]
This simplifies to:
\[
-70f + 4c + 200 = 0 \quad \text{(Equation 5)}
\]
### Step 7: Substitute \( g \) in Equation 2
Now substitute \( g \) into Equation 2:
\[
2\left(-\frac{3}{4}f\right) + 14f + c + 50 = 0
\]
This simplifies to:
\[
-\frac{3}{2}f + 14f + c + 50 = 0
\]
Multiplying through by 2 to eliminate the fraction:
\[
-3f + 28f + 2c + 100 = 0
\]
This simplifies to:
\[
25f + 2c + 100 = 0 \quad \text{(Equation 6)}
\]
### Step 8: Solve for \( f \) and \( c \)
Now we have two equations (Equation 5 and Equation 6):
1. \( -70f + 4c + 200 = 0 \)
2. \( 25f + 2c + 100 = 0 \)
We can solve these equations simultaneously. Let's multiply Equation 6 by 2:
\[
50f + 4c + 200 = 0
\]
Now subtract Equation 5 from this:
\[
(50f + 4c + 200) - (-70f + 4c + 200) = 0
\]
This simplifies to:
\[
120f = 0 \implies f = 0
\]
### Step 9: Substitute \( f \) back to find \( g \) and \( c \)
Substituting \( f = 0 \) into \( g = -\frac{3}{4}f \):
\[
g = 0
\]
Now substitute \( f = 0 \) into Equation 5 to find \( c \):
\[
-70(0) + 4c + 200 = 0 \implies 4c + 200 = 0 \implies c = -50
\]
### Final Equation of the Circle
Now we have \( g = 0 \), \( f = 0 \), and \( c = -50 \). The equation of the circle is:
\[
x^2 + y^2 + 0 + 0 - 50 = 0
\]
Thus, the final equation of the circle is:
\[
x^2 + y^2 = 50
\]
