1. **State the problem:** Write the standard form equation of the circle given by the equation $137 + 6y = -y^2 - x^2 - 24x$. 2. **Rewrite the equation:** Move all terms to one side to set the equation equal to zero: $$137 + 6y + y^2 + x^2 + 24x = 0$$ which simplifies to $$x^2 + 24x + y^2 + 6y + 137 = 0$$ 3. **Group x and y terms:** $$ (x^2 + 24x) + (y^2 + 6y) = -137 $$ 4. **Complete the square for x:** Take half of 24, which is 12, and square it: $12^2 = 144$. Add and subtract 144 inside the equation: $$ (x^2 + 24x + 144) - 144 + (y^2 + 6y) = -137 $$ 5. **Complete the square for y:** Take half of 6, which is 3, and square it: $3^2 = 9$. Add and subtract 9: $$ (x^2 + 24x + 144) - 144 + (y^2 + 6y + 9) - 9 = -137 $$ 6. **Rewrite as perfect squares:** $$ (x + 12)^2 - 144 + (y + 3)^2 - 9 = -137 $$ 7. **Combine constants:** $$ (x + 12)^2 + (y + 3)^2 = -137 + 144 + 9 $$ $$ (x + 12)^2 + (y + 3)^2 = 16 $$ 8. **Final standard form:** $$\boxed{(x + 12)^2 + (y + 3)^2 = 16}$$ This represents a circle with center $(-12, -3)$ and radius $4$ (since $\sqrt{16} = 4$).