1. **State the problem:** Find the equation of a circle that touches the coordinate axes at points (1,0) and (0,1). 2. **Understand the problem:** A circle touching the coordinate axes means it is tangent to both the x-axis and y-axis. The points of tangency are given as (1,0) and (0,1). 3. **Formula and properties:** The general equation of a circle is $$ (x - h)^2 + (y - k)^2 = r^2 $$ where $(h,k)$ is the center and $r$ is the radius. 4. Since the circle touches the x-axis at (1,0), the distance from the center to the x-axis must be equal to the radius. Similarly, since it touches the y-axis at (0,1), the distance from the center to the y-axis must also be equal to the radius. 5. The distance from the center $(h,k)$ to the x-axis is $|k|$ and to the y-axis is $|h|$. Since the circle touches both axes, we have: $$ r = |h| = |k| $$ 6. The points of tangency lie on the circle, so substituting (1,0) into the circle equation: $$ (1 - h)^2 + (0 - k)^2 = r^2 $$ 7. Substituting (0,1) into the circle equation: $$ (0 - h)^2 + (1 - k)^2 = r^2 $$ 8. Since $r = |h| = |k|$, assume $h = k = r$ (both positive because points of tangency are in the first quadrant). 9. Substitute $h = k = r$ into the equations: $$ (1 - r)^2 + (0 - r)^2 = r^2 $$ $$ (0 - r)^2 + (1 - r)^2 = r^2 $$ Both equations are the same, so solve one: 10. Expand and simplify: $$ (1 - r)^2 + r^2 = r^2 $$ $$ (1 - r)^2 = 0 $$ 11. Solve for $r$: $$ 1 - 2r + r^2 = 0 $$ $$ r^2 - 2r + 1 = 0 $$ $$ (r - 1)^2 = 0 $$ $$ r = 1 $$ 12. Therefore, the center is at $(1,1)$ and radius $r=1$. 13. The equation of the circle is: $$ (x - 1)^2 + (y - 1)^2 = 1^2 $$ $$ (x - 1)^2 + (y - 1)^2 = 1 $$ **Final answer:** The equation of the circle is $$ (x - 1)^2 + (y - 1)^2 = 1 $$.