To find the equation of the circle passing through the points (3, 4), (3, 2), and (1, 4), we can use the general form of the equation of a circle:
\[
x^2 + y^2 + Dx + Ey + F = 0
\]
where \(D\), \(E\), and \(F\) are constants that we need to determine. Since the circle passes through the three given points, we can substitute each point into the equation to create a system of equations.
### Step 1: Substitute the first point (3, 4)
Substituting \(x = 3\) and \(y = 4\):
\[
3^2 + 4^2 + 3D + 4E + F = 0
\]
This simplifies to:
\[
9 + 16 + 3D + 4E + F = 0 \implies 3D + 4E + F = -25 \quad \text{(Equation 1)}
\]
### Step 2: Substitute the second point (3, 2)
Substituting \(x = 3\) and \(y = 2\):
\[
3^2 + 2^2 + 3D + 2E + F = 0
\]
This simplifies to:
\[
9 + 4 + 3D + 2E + F = 0 \implies 3D + 2E + F = -13 \quad \text{(Equation 2)}
\]
### Step 3: Substitute the third point (1, 4)
Substituting \(x = 1\) and \(y = 4\):
\[
1^2 + 4^2 + 1D + 4E + F = 0
\]
This simplifies to:
\[
1 + 16 + D + 4E + F = 0 \implies D + 4E + F = -17 \quad \text{(Equation 3)}
\]
### Step 4: Solve the system of equations
Now we have the following system of equations:
1. \(3D + 4E + F = -25\) (Equation 1)
2. \(3D + 2E + F = -13\) (Equation 2)
3. \(D + 4E + F = -17\) (Equation 3)
We can eliminate \(F\) from the equations. Subtract Equation 2 from Equation 1:
\[
(3D + 4E + F) - (3D + 2E + F) = -25 + 13
\]
\[
2E = -12 \implies E = -6
\]
Now substitute \(E = -6\) into Equation 3:
\[
D + 4(-6) + F = -17 \implies D - 24 + F = -17 \implies D + F = 7 \quad \text{(Equation 4)}
\]
Now substitute \(E = -6\) into Equation 1:
\[
3D + 4(-6) + F = -25 \implies 3D - 24 + F = -25 \implies 3D + F = -1 \quad \text{(Equation 5)}
\]
### Step 5: Solve Equations 4 and 5
Now we have:
1. \(D + F = 7\) (Equation 4)
2. \(3D + F = -1\) (Equation 5)
Subtract Equation 4 from Equation 5:
\[
(3D + F) - (D + F) = -1 - 7
\]
\[
2D = -8 \implies D = -4
\]
Now substitute \(D = -4\) into Equation 4:
\[
-4 + F = 7 \implies F = 11
\]
### Step 6: Finalize the equation
Now we have \(D = -4\), \(E = -6\), and \(F = 11\). The equation of the circle is:
\[
x^2 + y^2 - 4x - 6y + 11 = 0
\]
### Final Answer
The equation of the circle passing through the points (3, 4), (3, 2), and (1, 4) is:
\[
x^2 + y^2 - 4x - 6y + 11 = 0
\]
