rational points in $\Bbb{P}^2(\Bbb{F}_q)$ I'm interested in the number of solutions $x^2＋y^2＝1$ in  $\Bbb{F}_q$.
I searched this site also, and by combining the information obtained there, I made the following answer.
The curve $x^2+y^2=1$ has genus zero, so its projective version $X^2+Y^2+Z^2=0$ always has exactly $q+1$ points in $\Bbb{P}^2(\Bbb{F}_q)$. Two of those points will lie on the line at infinity when $-1$ has a square root in $\Bbb{F}_q$.So,when $q≡1mod4$, $q$ー1 is the answer, and when $q≡3mod4$, $q＋1$ is the answer.
I have two questions.
① Why genus zero deduces the curve has $q＋1$ points?
②Why infinity consists of two points?
 A: Note that your answer doesn't address what happens when $q$ is even. This case degenerates because $x^2 + y^2 = (x + y)^2$ so the equation reduces to $x + y = 1$ which, of course, has $q$ solutions.
Q1: Abstractly, a smooth projective curve of genus $0$ with a rational point is isomorphic to $\mathbb{P}^1$. Concretely, in this case we can just write down such an isomorphism: it's given by
$$(A : B) \mapsto (A^2 - B^2 : 2AB : A^2 + B^2)$$
which is just the familiar formula for Pythagorean triples, and is the homogeneous version of the map given by Jyrki in the comments. It's a good exercise to see how this map arises by intersecting the curve $C$ with a line passing through $(1 : 0 : 1)$ in $\mathbb{P}^2$ (visualizing the real affine case of a line passing through $(1, 0)$ intersecting a circle may help).
Q2: As Jyrki says in the comments, the points at infinity are the points satisfying $X^2 + Y^2 = 0$. If $q \equiv 1 \bmod 4$ or $q$ is even then $-1$ has a square root so there are two points at infinity $(1 : \pm i : 0)$. If $q \equiv 3 \bmod 4$ then $-1$ is not a square root so there are no points at infinity. And if $q$ is even then $X^2 + Y^2 = (X + Y)^2$ so there is one point at infinity $(1 : 1 : 0)$.

An alternate way to produce this count is to write $x^2 + y^2$ as the norm $N(x + iy)$ where $x + iy \in \mathbb{F}_q[i]/(i^2 + 1)$. The isomorphism type of this quotient depends on how $t^2 + 1$ factors over $\mathbb{F}_q[t]$.
If $q \equiv 3 \bmod 4$ then you can check that $t^2 + 1$ is irreducible so $\mathbb{F}_q[i] \cong \mathbb{F}_{q^2}$. You can also check that the norm $N : \mathbb{F}_{q^2} \to \mathbb{F}_q$ is surjective, so its kernel has size $\frac{q^2 - 1}{q - 1} = q + 1$.
If $q \equiv 1 \bmod 4$ you can check that $t^2 + 1$ is reducible so $\mathbb{F}_q[i] \cong \mathbb{F}_q^2$. The norm is again surjective, so its kernel has size $\frac{(q - 1)^2}{q - 1} = q - 1$.
If $q$ is even $t^2 + 1 = (t + 1)^2$ so $\mathbb{F}_q[i] \cong \mathbb{F}_q[\varepsilon]/\varepsilon^2$ has nilpotents. The norm is again surjective, so its kernel has size $\frac{q(q - 1)}{q - 1} = q$.
