The image of a line through an inversion

Question

I am going to start by saying that geometry is not my strong suit, but I am taking a course on analytic geometry where I learnt about inversions and there is this question that bugs me.
The following is a problem that I encountered while working through a problem sheet:
Let $\mathcal{I}_{O, k}$ be the inversion of centre $O(0,0)$ and ratio $k=4$. Find $\mathcal{I}_{O, k}$(d), where $d:x-y+6=0$.
My professor solved this by using the fact that the image of a line not passing through the pole of inversion is a circle that passes through the pole and whose diametre also passes through the pole and is perpendicular to $d$. This solution is fine, but I wondered if it is not possible to solve this in a shorter way. Here is what I came up with.
The equation of our inversion is $\mathcal{I}_{O,k}: \begin{cases} x'=\frac{4x}{x^2+y^2} \\ y'=\frac{4y}{x^2+y^2} \end{cases}$. Since $\mathcal{I}_{O,k}\circ \mathcal{I}_{O, k}=\operatorname{id}_{\mathcal{E}_2\setminus\{O\}}$, we get that $\begin{cases} x=\frac{4x'}{x'^2+y'^2} \\ y=\frac{4y'}{x'^2+y'^2} \end{cases}$.
Let $\mathcal{C}:=\mathcal{I}_{O,k}(d)$. Now we get that $\mathcal{C}:\frac{4x'}{x'^2+y'^2}-\frac{4y'}{x'^2+y'^2}+6=0$ and after clearing the denominators we obtain that $\mathcal{C}:\left(x'+\frac{1}{3}\right)^2+\left(y'-\frac{1}{3}\right)^2=\frac{2}{9}$. This is precisely the equation that my professor obtained through that geometric procedure I presented above. The inspiration for my approach was the way we used to find, say, the image of a line through a rotation. I can't find any mistakes in my solution (maybe just the fact that I am kind of dividing by zero before clearing the denominators, but I am sure that this can be tweaked somehow), but I am not sure whether it is correct or not. So, what do you guys think?
EDIT: I think that a way to avoid that division by zero is to write $\mathcal{C}\setminus\{O\}:\frac{4x'}{x'^2+y'^2}-\frac{4y'}{x'^2+y'^2}+6=0$ instead of $\mathcal{C}:\frac{4x'}{x'^2+y'^2}-\frac{4y'}{x'^2+y'^2}+6=0$ (actually, I think that this is really the correct way, since the inversion is defined on $\mathcal{E}_2\setminus\{O\}$ and takes values in $\mathcal{E}_2\setminus\{O\}$). Then we would arrive at $\mathcal{C}\setminus\{O\}: \left(x'+\frac{1}{3}\right)^2+\left(y'-\frac{1}{3}\right)^2=\frac{2}{9}$ and since $O$ also verifies this equation we may write that $\mathcal{C}: \left(x'+\frac{1}{3}\right)^2+\left(y'-\frac{1}{3}\right)^2=\frac{2}{9}$.