For what value of $c$ will the circle with equation $x^2 - 10x + y^2 + 6y + c = 0$ have a radius of length 1?
I recognize that this is the equation of a circle in standard form, where the center is $(h, k)$ and the radius is $r$, and it can be written as $(x - h)^2 + (y - k)^2 = r^2$. I want to rewrite the given equation in this form, so I need to complete the square for both the $x$ and $y$ terms. To complete the square for the $x$ terms, I need to add and subtract $(10/2)^2 = 25$ inside the parentheses, and do the same outside to keep the equation balanced. To complete the square for the $y$ terms, I need to add and subtract $(6/2)^2 = 9$ inside the parentheses, and do the same outside. This gives me $(x^2 - 10x + 25) - 25 + (y^2 + 6y + 9) - 9 + c = 0$. Simplifying, I get $(x - 5)^2 + (y + 3)^2 - 34 + c = 0$. Now I can compare this to the standard form and see that the center is $(5, -3)$ and the radius is $\sqrt{34 - c}$. I want the radius to be 1, so I need to solve for $c$ such that $\sqrt{34 - c} = 1$. Squaring both sides, I get $34 - c = 1$, or $c = 33$. 