Suppose we have a action of cyclic group of order $n$ on $\mathbb{P}^2$ by $k\to \{(x,y,z)\to(x,e^{2\pi ik/n}y,z)\}$. It has a fixed point and a fixed line.
Is $\mathbb{P}^2/C_n$ a variety/scheme? If it is, can we write out its defining equations? And how do we describe the quotient singularities?
It is a (normal) projective variety, the weighted projective space $\mathbb{P}(1,n,1)$, which has one singular point, a so called cyclic singularity of type $\frac{1}{n}(1,1)$, corresponding to the fixed point $(0:1:0)$. This can be seen in greater generality, see for example the survey article Weighted projective varieties by I. Dolgachev.
In this special case, consider $\mathbb{P}^2$ as $\mathrm{Proj}(\mathbb{C}[x_0,x_1,x_2])$ with the standard grading. We have the action of the group $\mu_n$ of $n$-th roots of unity on $\mathbb{C}[x_0,x_1,x_2]$ (and $\mathbb{P}^2$ as in the question) via $\zeta.x_0 = x_0$, $\zeta.x_1 = \zeta x_1$ and $\zeta.x_2 = x_2$. It is easy to see that the invariant ring $\mathbb{C}[x_0,x_1,x_2]^{\mu_n}$ is generated by $x_0$, $x_2$ and $x_1^n$. That is, $\mathbb{C}[x_0,x_1,x_2]^{\mu_n}\subset\mathbb{C}[x_0,x_1,x_2]$ is the image of the injective homomorphism $\varphi\colon\mathbb{C}[y_0,y_1,y_2]\to\mathbb{C}[x_0,x_1,x_2]$ mapping $y_0$ to $x_0$, $y_1$ to $x_1^6$ and $y_2$ to $x_2$. In order to make this an homomorphism of graded $k$-algebras, we only have to give $\mathbb{C}[y_0,y_1,y_2]$ another grading, namely such that $y_0$ and $y_2$ are of degree $1$ as usual and letting $y_1$ be of degree $6$. We thus get an isomorphism $\mathrm{Proj}(\mathbb{C}[x_0,x_1,x_2]^{\mu_n})\to \mathrm{Proj}(\mathbb{C}[y_0,y_1,y_2])$. (The latter is by definition the weighted projective space $\mathbb{P}(1,n,1)$.)
Note that $\varphi$ induces an everywhere defined morphism $f := \varphi^*\colon\mathbb{P}^2\to\mathrm{Proj}(\mathbb{C}[y_0,y_1,y_2])$. In fact, it is well defined exactly on the $\mathfrak{p}\in\mathrm{Proj}(\mathbb{C}[x_0,x_1,x_2])$ that do not contain the image of the irrelevant ideal, i.e. such that $(x_0,x_1^6,x_2)\not\subset\mathfrak{p}$. But this is true for all of them, because a prime ideal $\mathfrak{p}$ containing $x_0,x_1^6$ and $x_2$ yet contains $(x_0,x_1,x_2)$, thus $\mathfrak{p}\not\in\mathrm{Proj}(\mathbb{C}[x_0,x_1,x_2])$. This should actually be the universal geometric quotient in the sense of Mumford, but this is probably too abstract, so let's get more concrete.
People will tell you that $\mathbb{P}(1,n,1)$ is the contraction of the exceptional section of the $n$-th Hirzebruch surface, or the projective cone of the Veronese $v_n(\mathbb{P}^1)$. The latter becomes very simple for $n=2$: consider the map $p\colon\mathbb{C}^3\setminus 0\to\mathbb{P}^3$ defined by $(x_0,x_1,x_2)\mapsto(x_0^2:x_1:x_0x_2:x_2^2)$. For any $\lambda\in\mathbb{C}^\times$, $p(\lambda x_0,\lambda^2 x_1,\lambda x_2) = (x_0^2:x_1:x_0x_2:x_2^2)$ and $p(y_0,y_1,y_2) = p(x_0,x_1,x_2)$ if and only if $(y_0,y_1,y_2)=(\lambda x_0,\lambda^2 x_1,\lambda x_2)$ for some $\lambda\in\mathbb{C}^\times$. This is another defining property of $\mathbb{P}(1,2,1)$, i.e. being the quotient of $\mathbb{C}^3\setminus 0$ by the $\mathbb{C}^\times$-action $\lambda.(x_0,x_1,x_2) = (\lambda x_0,\lambda^2 x_1,\lambda x_2)$. Consequently, $\mathbb{P}(1,2,1)$ is the image of $p$, which is just the quadratic cone $\{(w:x:y:z)\in\mathbb{P}^3|\,wz=y^2\}$. Its only singular point is $(0:1:0:0)$, whose $p$-fibre is the line in $\mathbb{C}^3\setminus 0$ generated by $(0,1,0)$. In view of the original question, this corresponds to the fixed point $(0:1:0)\in\mathbb{P}^2$. Analogously, for any $n\geq 2$, the map $\mathbb{C}^3\setminus 0\to \mathbb{P}^{n+1}$, $(x,y,z)\mapsto (x^n,x^{n-1}z,x^{n-2}z^2,\dots,z^n,y)$ does the same job.
