1. **State the problem:** We need to find the equation of a circle that passes through the points $(0,0)$ and $(4,2)$, and whose center lies on the line $x + y = 1$. 2. **Formula for a circle:** The equation of a circle with center $(h,k)$ and radius $r$ is: $$ (x - h)^2 + (y - k)^2 = r^2 $$ 3. **Using the condition that the circle passes through $(0,0)$:** Substitute $x=0$, $y=0$: $$ (0 - h)^2 + (0 - k)^2 = r^2 $$ $$ h^2 + k^2 = r^2 $$ 4. **Using the condition that the circle passes through $(4,2)$:** Substitute $x=4$, $y=2$: $$ (4 - h)^2 + (2 - k)^2 = r^2 $$ 5. **Equate the two expressions for $r^2$:** $$ h^2 + k^2 = (4 - h)^2 + (2 - k)^2 $$ 6. **Expand the right side:** $$ h^2 + k^2 = (4 - h)^2 + (2 - k)^2 = (16 - 8h + h^2) + (4 - 4k + k^2) $$ $$ h^2 + k^2 = 16 - 8h + h^2 + 4 - 4k + k^2 $$ 7. **Simplify by canceling $h^2$ and $k^2$ on both sides:** $$ \cancel{h^2} + \cancel{k^2} = 16 - 8h + \cancel{h^2} + 4 - 4k + \cancel{k^2} $$ $$ 0 = 20 - 8h - 4k $$ 8. **Rewrite the equation:** $$ 8h + 4k = 20 $$ 9. **Divide both sides by 4:** $$ \frac{8h}{4} + \frac{4k}{4} = \frac{20}{4} $$ $$ 2h + k = 5 $$ 10. **Use the condition that the center lies on the line $x + y = 1$:** $$ h + k = 1 $$ 11. **Solve the system of equations:** $$ \begin{cases} 2h + k = 5 \\ h + k = 1 \end{cases} $$ 12. **Subtract the second equation from the first:** $$ (2h + k) - (h + k) = 5 - 1 $$ $$ 2h + k - h - k = 4 $$ $$ h = 4 $$ 13. **Substitute $h=4$ into $h + k = 1$:** $$ 4 + k = 1 $$ $$ k = 1 - 4 = -3 $$ 14. **Find the radius squared $r^2$ using $h^2 + k^2 = r^2$:** $$ r^2 = 4^2 + (-3)^2 = 16 + 9 = 25 $$ 15. **Write the equation of the circle:** $$ (x - 4)^2 + (y + 3)^2 = 25 $$ **Final answer:** The equation of the circle is $$ (x - 4)^2 + (y + 3)^2 = 25 $$