To find the equation of a circle passing through the points $(2, 3)$, $(3, 4)$, and $(-1, 2)$, we can use the general equation of a circle:

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

where $D$, $E$, and $F$ are constants we need to determine.

Step 1: Set up the system of equations

Substitute the three points into the general equation of the circle:

For $(2, 3)$: $(2)^2 + (3)^2 + 2D + 3E + F = 0 \quad \Rightarrow \quad 4 + 9 + 2D + 3E + F = 0 \quad \Rightarrow \quad 2D + 3E + F = -13$

For $(3, 4)$: $(3)^2 + (4)^2 + 3D + 4E + F = 0 \quad \Rightarrow \quad 9 + 16 + 3D + 4E + F = 0 \quad \Rightarrow \quad 3D + 4E + F = -25$

For $(-1, 2)$: $(-1)^2 + (2)^2 + (-1)D + 2E + F = 0 \quad \Rightarrow \quad 1 + 4 - D + 2E + F = 0 \quad \Rightarrow \quad -D + 2E + F = -5$

Step 2: Solve the system of equations

Now we have the following system of equations:

1) & \quad 2D + 3E + F = -13 \\ 2) & \quad 3D + 4E + F = -25 \\ 3) & \quad -D + 2E + F = -5 \end{aligned}$$ Let's solve this system step by step. We can solve it using elimination or substitution. I'll solve it now. The solution to the system of equations is: $$D = 2, \quad E = -14, \quad F = 25$$ ### Step 3: Write the equation of the circle Substitute $$D = 2$$, $$E = -14$$, and $$F = 25$$ into the general equation of the circle: $$x^2 + y^2 + 2x - 14y + 25 = 0$$ Thus, the equation of the circle passing through the points $$(2, 3)$$, $$(3, 4)$$, and $$(-1, 2)$$ is: $$x^2 + y^2 + 2x - 14y + 25 = 0$$ Would you like further details on any step? Here are some related questions you might find interesting: 1. How do you derive the equation of a circle from its center and radius? 2. Can a circle pass through more than three points? 3. What are the properties of the general equation of a circle? 4. How can we verify if a point lies on a given circle? 5. What happens if the three points are collinear when finding a circle? **Tip:** You can check if the points are collinear by ensuring the determinant of a matrix with these points is non-zero.