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 a circle that passes through three points \( A(2, 3) \), \( B(6, 1) \), and \( C(4, -3) \), we can use the general equation of a circle:

\[
x^2 + y^2 + Dx + Ey + F = 0
\]

Here, \( D \), \( E \), and \( F \) are constants that we need to determine.

Given that the circle passes through points \( A(2, 3) \), \( B(6, 1) \), and \( C(4, -3) \), we substitute these points into the general equation to form a system of linear equations.

### Step 1: Substitute each point into the circle equation.

For point \( A(2, 3) \):

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

For point \( B(6, 1) \):

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

For point \( C(4, -3) \):

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

### Step 2: Solve the system of equations.

We now solve the following system of linear equations:

\[
\begin{aligned}
2D + 3E + F &= -13 &\quad \text{(Equation 1)} \\
6D + E + F &= -37 &\quad \text{(Equation 2)} \\
4D - 3E + F &= -25 &\quad \text{(Equation 3)}
\end{aligned}
\]

By solving this system, we can find the values of \( D \), \( E \), and \( F \), which will give us the specific equation of the circle.

### Step 3: Subtract Equation 1 from Equation 2:

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

### Step 4: Subtract Equation 3 from Equation 1:

\[
(4D - 3E + F) - (2D + 3E + F) = -25 - (-13)
\]
\[
2D - 6E = -12
\]
\[
D - 3E = -6 \quad \text{(Equation 5)}
\]

### Step 5: Solve Equations 4 and 5.

We now solve the system of equations:

\[
\begin{aligned}
2D - E &= -12 &\quad \text{(Equation 4)} \\
D - 3E &= -6 &\quad \text{(Equation 5)}
\end{aligned}
\]

From Equation 5:

\[
D = 3E - 6
\]

Substitute into Equation 4:

\[
2(3E - 6) - E = -12
\]
\[
6E - 12 - E = -12
\]
\[
5E = 0
\]
\[
E = 0
\]

Substituting \( E = 0 \) into Equation 5:

\[
D - 3(0) = -6
\]
\[
D = -6
\]

Finally, substitute \( D = -6 \) and \( E = 0 \) into Equation 1:

\[
2(-6) + 3(0) + F = -13
\]
\[
-12 + F = -13
\]
\[
F = -1
\]

### Final Equation:

Substituting \( D = -6 \), \( E = 0 \), and \( F = -1 \) back into the general equation:

\[
x^2 + y^2 - 6x - 1 = 0
\]

Thus, the equation of the circle is:

\[
x^2 + y^2 - 6x - 1 = 0
\]

Would you like to dive deeper into any of these steps?

Here are 5 related questions:
1. How can you find the center and radius of the circle from its general equation?
2. What are the properties of a circle in the coordinate plane?
3. How do you determine if a point lies inside, outside, or on the circle?
4. Can you derive the equation of a circle given its center and radius?
5. How does the general form of a circle equation relate to its standard form?

**Tip:** When solving systems of equations, reducing the number of variables step-by-step by elimination or substitution helps simplify the solution process.

Discover how to determine the general equation of a circle passing through specific points using systems of linear equations. This problem involves analytic geometry and circle equations, suitable for high school students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation