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:

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.