$y = −x + \sqrt{2}$, $y = −x − \sqrt{2}$, $y = x + \sqrt{2}$, and $y = x − \sqrt{2}$. These equations determine lines, which in turn bound a diamond shaped region in the plane. Construct a diamond shaped region in which the circle of radius $1$ centered at $(−5, −4)$ sits tangentially. Use the techniques of this section to help. (Order your answers from smallest to largest slope, then from smallest to largest $y$-intercept. Use $x$ as your variable.) so the answers I got were:
$\sqrt{2-(x+5)^2} +4$,
$\sqrt{2-(-x+5)^2}+4$,
$\sqrt{2-(x+5)^2}+4$
$\sqrt{2-(-x+5)^2}+4$
I dont even know if this is right but this is what I got when I tried to include the shifting from the $(0,0)$ center to the $(-5,-4)$ center.
I'm pretty sure all you need to do is translate each original solution by $(-5, -4)$ (i.e. to the left by $5$ and down by $4$). If you have a function, $f(x)$, and you want to translate it by $(x_0, y_0)$ then you do the following: $g(x) = f(x - x_0) + y_0$ where $g(x)$ is the translated function. So that's it--all you need to do is translate each of those functions. This means adding $5$ to $x$ (since you should subtract $-5$) and subtracting $4$ from that:
Here is one of them (note the necessary parentheses):
$$ y_1 = -x + \sqrt{2} \rightarrow \text{ translate to the left by } 5 \\ y_{\leftarrow 5} = -(x + 5) + \sqrt{2} \rightarrow \text{ translate down by } 4 \\ y_{\leftarrow 5, \downarrow 4} = \left(-x - 5 + \sqrt{2}\right) - 4 = -x + \sqrt{2} - 9 $$