Answer
$(x-5)^2-(y-5)^2=1$
hyperbola center $C(5,5)$, foci $F(5\pm\sqrt 2, 5)$, vertices $V_1(4,5)$ and $V_2(6,5)$
asymptotes: $y=x$ and $y=-x+10$
See graph.
Work Step by Step
Step 1. Form square with variables: $x^2-10x+25-y^2+10y-25=1$ or $(x-5)^2-(y-5)^2=1$
Step 2. The above equation represents a hyperbola with a center at $C(5,5)$
Step 3. With $a=1,b=1$, we have $c=\sqrt {1+1}=\sqrt 2$ and foci at $(5\pm\sqrt 2, 5)$
Step 4. The vertices are at $(5\pm1,5)$ or $V_1(4,5)$ and $V_2(6,5)$
Step 5. The original asymptotes are $y=\pm x$ and the new asymptotes are $y-5=\pm (x-5)$ or $y=\pm (x-5)+5$ which gives $y=x$ and $y=-x+10$
Step 6. See graph.
