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

$Ax^2 + Ay^2 + Dx + Ey + F = 0$

Since it’s a circle, the coefficients of $x^2$ and $y^2$ are equal ($A$ and $C$), and we’re instructed to set $A = C = 1$. Therefore, the equation simplifies to:

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

Step 1: Substitute the Points into the Equation

We’ll substitute each of the points into this equation to create a system of equations.

1. Using $(-1,1)$: $(-1)^2 + (1)^2 + D(-1) + E(1) + F = 0$ $1 + 1 - D + E + F = 0$ $2 - D + E + F = 0$

2. Using $(2,3)$: $(2)^2 + (3)^2 + D(2) + E(3) + F = 0$ $4 + 9 + 2D + 3E + F = 0$ $13 + 2D + 3E + F = 0$

3. Using $(4,2)$: $(4)^2 + (2)^2 + D(4) + E(2) + F = 0$ $16 + 4 + 4D + 2E + F = 0$ $20 + 4D + 2E + F = 0$

Step 2: Set Up the System of Equations

Now we have the following system of linear equations: