For what value of $c$ will the circle with equation $x^2 - 10x + y^2 + 10y + c = 0$ have a radius of length 1?
We can complete the square to get \[\begin{aligned} x^2-10x+y^2+10y+c &= 0 \\ x^2-10x+25+y^2+10y+25+c &= 50 \\ (x-5)^2+(y+5)^2 &= 49-c. \end{aligned}\]If the radius is 1, then the right side must be equal to 1, so $49-c=1$. Therefore, $c=\boxed{48}$.
To find the value of $c$ that will give the circle a radius of length $1$, we need to first complete the square for both $x$ and $y$ terms.
Starting with the equation $x^2 - 10x + y^2 + 10y + c = 0$, we can complete the square for $x$ by adding $\left(\frac{-10}{2}\right)^2 = 25$ to both sides of the equation:
$(x^2 - 10x) + 25 + y^2 + 10y + c = 25$
$(x^2 - 10x + 25) + y^2 + 10y + c = 25$
$(x - 5)^2 + y^2 + 10y + c = 25$
Similarly, we complete the square for $y$ by adding $\left(\frac{10}{2}\right)^2 = 25$ to both sides:
$(x - 5)^2 + (y^2 + 10y) + 25 + c = 25 + 25$
$(x - 5)^2 + (y^2 + 10y + 25) + c = 50$
$(x - 5)^2 + (y + 5)^2 + c = 50$
Now we can see that the equation of the circle is in the form $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center of the circle and $r$ is the radius. So, comparing our equation to this form, we have:
$(x - 5)^2 + (y + 5)^2 = \sqrt{50}$
Since the radius of the circle is $1$, we can rewrite this equation as:
$(x - 5)^2 + (y + 5)^2 = 1^2$
Expanding the equation and comparing coefficients, we have:
$(x^2 - 10x + 25) + (y^2 + 10y + 25) = 1$
$x^2 - 10x + y^2 + 10y + 50 = 1$
Comparing this to the given equation, we can see that $c = 50 - 1 = 49$.
Therefore, the value of $c$ that will give the circle a radius of length $1$ is $c = \boxed{49}$.