CONTEXT: First year university mathematics course question
Given an equation for displacement in metres is $R(x,y,z)=\frac{25z^2+30}{x^2+y^2}+40$, where $(x,y,z)$ represents coordinates in three-dimensional space and a point $M(3,1,2)$, find an expression for the surface which has the same displacement as M.
I found the displacement at M after substituting it into the equation to be $53$ metres, but am unsure of where to go from here.
Does this mean I let $R(x,y,z)=53$ which simplifies to $0=\frac{25z^2+30}{x^2+y^2}-13$, and then just set the LHS to be $R_1(x,y,z)$ so I have $R_1(x,y,z)=\frac{25z^2+30}{x^2+y^2}-13$? Then, my expression for the surface is $\frac{25z^2+30}{x^2+y^2}-13$. I have a feeling this is incorrect but don't know how to trouble shoot it.
Any guidance would be greatly appreciated!