Describe the level curve $f(x,y)=2$ for $f(x,y)=\sqrt{7(x+11)^2+7(y−12)^2}$ I am trying to do this homework problem that wants me to find the level curve for the given function. I am trying hard but just cant seem to grasp any of this level curve stuff. Here's my problem:
Describe the level curve $f(x,y)=C$ for
$C=2$, where $$f(x,y)=\sqrt{7(x+11)^2+7(y−12)^2}.$$
Circle with radius $r = \ldots$ and center $(x,y)=( \ldots,\ldots )$.
When I did the problem I found the radius to be $4$ and the center to be $(-11,12)$ but this was wrong and I just can't seem to find the right way to do this problem. Can anyone help me please?

With the given $f(x,y)$ and level $C = 2$, the equation of the level curve becomes:
$$\sqrt{7(x+11)^2+7(y-12)^2} = 2$$
Squaring yields:
$$7(x+11)^2+7(y-12)^2 = 4$$
You got the center right, but for the radius you need to be careful. You said $4$, probably based on the $4$ in the RHS. Note however that there are two problems with that. Compare with the standard equation of a circle centered in $(x_0,y_0)$ and with radius $r$:
$$\color{blue}{(x-x_0)^2+(y-y_0)^2 = r^2}$$
In standard form, the RHS is not the radius but the radius squared and furthermore, the $\color{red}{7}$'s in
$$\color{red}{7}(x+11)^2+\color{red}{7}(y-12)^2 = 4$$
are not in the standard form.
Get rid of the $7$'s by dividing both sides by $7$ and then take into account that the RHS is $r^2$ (and not $r$) to get the right value for the radius.
