Write the general form of a circle which passes through the points A(2,3)B(6,1)C(4-3)

To find the general form of the equation of a circle that passes through three points \( A(2, 3) \), \( B(6, 1) \), and \( C(4, -3) \), we follow these steps:

### Step 1: General Form of the Circle
The general equation of a circle is:
\[
x^2 + y^2 + Dx + Ey + F = 0
\]
where \( D \), \( E \), and \( F \) are constants.

### Step 2: Substitute the Points into the General Form
Substitute the coordinates of the points \( A \), \( B \), and \( C \) into the equation to get three equations.

For point \( A(2, 3) \):
\[
(2)^2 + (3)^2 + 2D + 3E + F = 0
\]
\[
4 + 9 + 2D + 3E + F = 0 \quad \text{(Equation 1)}
\]
\[
2D + 3E + F = -13
\]

For point \( B(6, 1) \):
\[
(6)^2 + (1)^2 + 6D + 1E + F = 0
\]
\[
36 + 1 + 6D + E + F = 0 \quad \text{(Equation 2)}
\]
\[
6D + E + F = -37
\]

For point \( C(4, -3) \):
\[
(4)^2 + (-3)^2 + 4D - 3E + F = 0
\]
\[
16 + 9 + 4D - 3E + F = 0 \quad \text{(Equation 3)}
\]
\[
4D - 3E + F = -25
\]

### Step 3: Solve the System of Equations
We now have the following system of linear equations:
\[
\begin{align*}
2D + 3E + F &= -13 \quad \text{(1)}\\
6D + E + F &= -37 \quad \text{(2)}\\
4D - 3E + F &= -25 \quad \text{(3)}
\end{align*}
\]

Subtract Equation 1 from Equation 2:
\[
(6D + E + F) - (2D + 3E + F) = -37 - (-13)
\]
\[
4D - 2E = -24
\]
\[
2D - E = -12 \quad \text{(4)}
\]

Subtract Equation 3 from Equation 2:
\[
(6D + E + F) - (4D - 3E + F) = -37 - (-25)
\]
\[
2D + 4E = -12 \quad \text{(5)}
\]

### Step 4: Solve for \( D \), \( E \), and \( F \)
From Equation 4:
\[
E = 2D + 12
\]

Substitute into Equation 5:
\[
2D + 4(2D + 12) = -12
\]
\[
2D + 8D + 48 = -12
\]
\[
10D = -60
\]
\[
D = -6
\]

Now, substitute \( D = -6 \) into \( E = 2D + 12 \):
\[
E = 2(-6) + 12 = 0
\]

Substitute \( D = -6 \) and \( E = 0 \) into Equation 1 to find \( F \):
\[
2(-6) + 3(0) + F = -13
\]
\[
-12 + F = -13
\]
\[
F = -1
\]

### Step 5: Write the Equation
Substituting \( D = -6 \), \( E = 0 \), and \( F = -1 \) into the general form:
\[
x^2 + y^2 - 6x - 1 = 0
\]
This is the general form of the circle passing through the points \( A(2, 3) \), \( B(6, 1) \), and \( C(4, -3) \).

Would you like to see more details or have any questions?

Here are 5 related questions:
1. How do you convert the general form of a circle into its standard form?
2. How can you find the center and radius of a circle from the general form?
3. What is the significance of the coefficients \( D \), \( E \), and \( F \) in the general form of a circle?
4. How do you find the equation of a circle given its center and radius?
5. How do you determine if three points lie on a circle?

**Tip:** When solving a system of equations, it's often helpful to eliminate variables step by step to simplify the problem.

Learn how to find the general form of a circle passing through points A(2,3), B(6,1), and C(4,-3) using analytical geometry and linear algebra techniques. Follow a step-by-step solution involving the general equation of a circle and solving systems of linear equations.

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