Completing the square gives us $(x - 5)^2 + (y + 3)^2 = 34 - c$. Since we want the radius to be 1, we must have $34 - c = 1^2$. It follows that $c = \boxed{33}$.
<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
For what value of $c$ will the circle with equation $x^2 - 10x + y^2 + 6y + c = 0$ have a radius of length 1?
Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
<|im_start|>assistant

user
33
For what value of $c$ will the circle with equation $x^2 - 10x + y^2 + 6y + c = 0$ have a radius of length 1?
test