To find the equation of the circle passing through the points $(-3, 6)$, $(-5, 2)$, and $(3, -6)$, we can use the general equation of a circle:

$x^2 + y^2 + Dx + Ey + F = 0$

where $D$, $E$, and $F$ are constants that need to be determined.

Step 1: Substituting the points into the equation

We substitute each point into the general equation:

1. For $(-3, 6)$: $(-3)^2 + (6)^2 + D(-3) + E(6) + F = 0$ $9 + 36 - 3D + 6E + F = 0$ \quad \text{(Equation 1)}$$
2. For $(-5, 2)$: $(-5)^2 + (2)^2 + D(-5) + E(2) + F = 0$ $25 + 4 - 5D + 2E + F = 0$ \quad \text{(Equation 2)}$$
3. For $(3, -6)$: $(3)^2 + (-6)^2 + D(3) + E(-6) + F = 0$ $9 + 36 + 3D - 6E + F = 0$ \quad \text{(Equation 3)}$$

Step 2: Solving the system of equations

We now solve the system of equations formed by:

1. $45 - 3D + 6E + F = 0$
2. $29 - 5D + 2E + F = 0$
3. $45 + 3D - 6E + F = 0$

Let's solve this step-by-step. I'll continue by eliminating $F$.The solution to the system of equations is:

• $D = -4$
• $E = -2$
• $F = -45$

Step 3: Writing the equation of the circle

Substituting these values into the general equation of a circle:

$x^2 + y^2 + Dx + Ey + F = 0$

we get:

$x^2 + y^2 - 4x - 2y - 45 = 0$

Thus, the equation of the circle is:

$x^2 + y^2 - 4x - 2y - 45 = 0$

Would you like a detailed explanation of any part of this solution? Here are some related questions for further exploration:

1. How do you find the radius of the circle given the equation?
2. Can we convert this equation into center-radius form?
3. How would we determine if a point lies inside the circle?
4. What is the geometric significance of the constants $D$ and $E$?
5. How do you generalize this method for any three points?

Tip: The general form of a circle equation $x^2 + y^2 + Dx + Ey + F = 0$ can be easily converted into center-radius form to identify the center and radius directly.