1. **Problem statement:** Find the equations of two circles that pass through the point $(6, 2)$, have centers on the line $x + y = 7$, and are tangent to the x-axis. 2. **Key facts and formulas:** - The general equation of a circle with center $(h, k)$ and radius $r$ is: $$ (x - h)^2 + (y - k)^2 = r^2 $$ - Since the circle is tangent to the x-axis, the radius $r$ equals the vertical distance from the center to the x-axis, so: $$ r = |k| $$ - The center $(h, k)$ lies on the line $x + y = 7$, so: $$ h + k = 7 $$ 3. **Express radius in terms of $k$:** $$ r = |k| $$ 4. **Use the point $(6, 2)$ on the circle:** Substitute $x=6$, $y=2$ into the circle equation: $$ (6 - h)^2 + (2 - k)^2 = r^2 = k^2 $$ 5. **Substitute $h = 7 - k$ from the line equation:** $$ (6 - (7 - k))^2 + (2 - k)^2 = k^2 $$ Simplify inside the first term: $$ (6 - 7 + k)^2 + (2 - k)^2 = k^2 $$ $$ (k - 1)^2 + (2 - k)^2 = k^2 $$ 6. **Expand the squares:** $$ (k - 1)^2 = k^2 - 2k + 1 $$ $$ (2 - k)^2 = k^2 - 4k + 4 $$ 7. **Sum and set equal to $k^2$:** $$ (k^2 - 2k + 1) + (k^2 - 4k + 4) = k^2 $$ $$ 2k^2 - 6k + 5 = k^2 $$ 8. **Bring all terms to one side:** $$ 2k^2 - 6k + 5 - k^2 = 0 $$ $$ k^2 - 6k + 5 = 0 $$ 9. **Solve quadratic equation for $k$:** $$ k = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 5}}{2} = \frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm \sqrt{16}}{2} $$ $$ k = \frac{6 \pm 4}{2} $$ 10. **Two solutions for $k$:** - $$ k = \frac{6 + 4}{2} = 5 $$ - $$ k = \frac{6 - 4}{2} = 1 $$ 11. **Find corresponding $h$ values:** $$ h = 7 - k $$ - For $k=5$, $h = 7 - 5 = 2$ - For $k=1$, $h = 7 - 1 = 6$ 12. **Find radii:** $$ r = |k| $$ - For $k=5$, $r=5$ - For $k=1$, $r=1$ 13. **Write the equations of the two circles:** - For center $(2, 5)$ and radius $5$: $$ (x - 2)^2 + (y - 5)^2 = 25 $$ - For center $(6, 1)$ and radius $1$: $$ (x - 6)^2 + (y - 1)^2 = 1 $$ **Final answer:** $$ (x - 2)^2 + (y - 5)^2 = 25 $$ $$ (x - 6)^2 + (y - 1)^2 = 1 $$