To find the equation of the circle passing through the points $(-2, -3)$, $(3, -3)$, and $(0, 0)$, we need to substitute these points into the general equation for a circle:

$a(x^2 + y^2) + bx + cy + d = 0$

Step 1: Substitute the points into the equation.

1. Substitute $(-2, -3)$: $a((-2)^2 + (-3)^2) + b(-2) + c(-3) + d = 0$ Simplifying: $a(4 + 9) - 2b - 3c + d = 0 \implies 13a - 2b - 3c + d = 0$

2. Substitute $(3, -3)$: $a(3^2 + (-3)^2) + 3b - 3c + d = 0$ Simplifying: $a(9 + 9) + 3b - 3c + d = 0 \implies 18a + 3b - 3c + d = 0$

3. Substitute $(0, 0)$: $a(0^2 + 0^2) + 0b + 0c + d = 0 \implies d = 0$

Since $d = 0$, the equation simplifies to:

$a(x^2 + y^2) + bx + cy = 0$

Step 2: Solve the system of linear equations.

The equations from substituting the points now are:

1. $13a - 2b - 3c = 0$
2. $18a + 3b - 3c = 0$

We solve these simultaneously.

Solve the first equation for $b$:

$b = \frac{13a - 3c}{2}$

Substitute $b$ into the second equation:

$18a + 3\left(\frac{13a - 3c}{2}\right) - 3c = 0$

Multiply the terms by 2 to eliminate the fraction:

$36a + 39a - 9c - 6c = 0 \implies 75a - 15c = 0 \implies 5a = c$

So, $c = 5a$.

Substitute $c = 5a$ back into $b = \frac{13a - 3c}{2}$:

$b = \frac{13a - 3(5a)}{2} = \frac{13a - 15a}{2} = \frac{-2a}{2} = -a$

Step 3: Write the equation of the circle.

Substitute $b = -a$, $c = 5a$, and $d = 0$ into the general form:

$a(x^2 + y^2) - ax + 5ay = 0$

Factor out $a$:

$a(x^2 + y^2 - x + 5y) = 0$

Since $a \neq 0$, we can divide by $a$ to get:

$x^2 + y^2 - x + 5y = 0$

This is the equation of the circle passing through the points $(-2, -3)$, $(3, -3)$, and $(0, 0)$.

Would you like more details or have any questions?

Relative Questions:

1. How do you find the center and radius of the circle given its equation?
2. Can the method used here be applied to find the circle passing through any three points?
3. What is the significance of the coefficients $b$ and $c$ in the general circle equation?
4. How does the equation change if the circle is centered at the origin?
5. How do you derive the standard form of a circle from the general form?

Tip:

When given three points, ensuring they are not collinear is crucial for a unique circle to exist that passes through them.