1. **State the problem:** Find the equation of a circle passing through points $(2,3)$ and $(-1,2)$ with its center on the line $2x - 3y + 1 = 0$. 2. **Formula and rules:** The general equation of a circle with center $(h,k)$ and radius $r$ is: $$ (x - h)^2 + (y - k)^2 = r^2 $$ The center $(h,k)$ lies on the line: $$ 2h - 3k + 1 = 0 $$ 3. **Use the condition that the circle passes through the points:** Substitute $(2,3)$: $$ (2 - h)^2 + (3 - k)^2 = r^2 $$ Substitute $(-1,2)$: $$ (-1 - h)^2 + (2 - k)^2 = r^2 $$ 4. **Set the two expressions for $r^2$ equal:** $$ (2 - h)^2 + (3 - k)^2 = (-1 - h)^2 + (2 - k)^2 $$ 5. **Expand and simplify:** $$ (2 - h)^2 = (h - 2)^2 = h^2 - 4h + 4 $$ $$ (3 - k)^2 = k^2 - 6k + 9 $$ $$ (-1 - h)^2 = (h + 1)^2 = h^2 + 2h + 1 $$ $$ (2 - k)^2 = k^2 - 4k + 4 $$ So, $$ h^2 - 4h + 4 + k^2 - 6k + 9 = h^2 + 2h + 1 + k^2 - 4k + 4 $$ 6. **Cancel $h^2$ and $k^2$ on both sides:** $$ -4h + 4 - 6k + 9 = 2h + 1 - 4k + 4 $$ 7. **Simplify further:** $$ -4h - 6k + 13 = 2h - 4k + 5 $$ 8. **Bring all terms to one side:** $$ -4h - 6k + 13 - 2h + 4k - 5 = 0 $$ $$ (-4h - 2h) + (-6k + 4k) + (13 - 5) = 0 $$ $$ -6h - 2k + 8 = 0 $$ 9. **Divide entire equation by -2 to simplify:** $$ \cancel{-6h} \cancel{-2k} \cancel{+8} = 0 \Rightarrow 3h + k - 4 = 0 $$ 10. **Now we have two linear equations for $h$ and $k$:** $$ 2h - 3k + 1 = 0 $$ $$ 3h + k - 4 = 0 $$ 11. **Solve the system:** From second equation: $$ k = 4 - 3h $$ Substitute into first: $$ 2h - 3(4 - 3h) + 1 = 0 $$ $$ 2h - 12 + 9h + 1 = 0 $$ $$ 11h - 11 = 0 $$ $$ 11h = 11 $$ $$ h = 1 $$ 12. **Find $k$:** $$ k = 4 - 3(1) = 4 - 3 = 1 $$ 13. **Center is $(1,1)$**. 14. **Find radius $r$ using point $(2,3)$:** $$ r^2 = (2 - 1)^2 + (3 - 1)^2 = 1^2 + 2^2 = 1 + 4 = 5 $$ 15. **Equation of the circle:** $$ (x - 1)^2 + (y - 1)^2 = 5 $$