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

Answer 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, $c=\boxed{48}$.