All the affine points on the curve I have calculated the affine points on the curve $x^2 + y^2 = 1 − 3x^2y^2$
over the field ${\mathbb{Z}}_{11}.$

Using $y^2 = \frac{1-x^2}{1+3x^2}$ I got the following points:

$(0,1),(0,10),(1,0),(2,2),(2,9),(9,2),(9,9),(10,0)$

Total comes to be 8 points. But what I know is that total points should come 11-1 = 10. 

So I am not getting other 2 points.
Well, this curve has 8 affine points and I would like to share the way to find them all.
First of all, we make a square table till 10 as we have  field ${\mathbb{Z}}_{11}.$  (so 11-1=10)
Below is the table(also calculate its mod 11 values) :
  ┌───────┬─────┬────┬────┬────┬────┬────┬────┬────┬─────┬─────┬────┬
  │   #   │  0  │  1 │  2 │  3 │ 4  │ 5  │ 6  │  7 │  8  │  9  │ 10 │
  └───────┴─────┴────┴────┴────┴────┴────┴────┴────┴─────┴─────┴────┴    
  │Square │  0  │  1 │  4 │  9 │ 16 │ 25 │ 36 │ 49 │  64 │  81 │ 100│
  └───────┴─────┴────┴────┴────┴────┴────┴────┴────┴─────┴─────┴────┴    
  │Mod 11 │  0  │  1 │  4 │  9 │ 5  │ 3  │ 3  │  5 │  9  │  4  │  1 │
  └───────┴─────┴────┴────┴────┴────┴────┴────┴────┴─────┴─────┴────┴    

Re-arranging the equation so that we have $y$ on the left-hand side
$y^2 = \frac{1-x^2}{1+3x^2}$
Now we start putting form $x = 0$  until $x = 10$  in the above equation and get the value of $y$ each time. If the value of $y$ is one of the complete squares mod 11 i.e row 3rd of the table then the corresponding value in row 1 is the $y$-coordinate.
$y$ is calculated as mod 11. Even if the value of $y$ is in fractions.
For $x=0$ we get $y^=1$ which exists in row 3rd i.e it is a complete square. Now value 1 in 3rd row corresponds to value $1$ and $10$ in row 1. So we get 2 points here i.e. $(0,1)$ and $(0,10)$
For $x=1$ we get $y^2=0$  this also exists in 3rd row. So corresponding values in 1st row are 0. So here we have only 1 point $(1,0)$
For $x=2$ we get $y^2=4$ this also exists in the 3rd row. So corresponding values in 1st row are 2 and 9. So here we have 2 points i.e. $(2,2)$ and $(2,9)$
For $x=3$ we get $y^2=6$  which does not exist in the 3rd row of the table. So no corresponding value of $y$ will be there, hence no point will exist at $x=3$
For $x=4$ we get $y^2=8$   Not in the table (not a complete square) hence no points here.
For $x=5$ we get $y^2=2$   Not in the table (not a complete square) hence no points here.
For $x=6$ we get $y^2=2$   Not in the table (not a complete square) hence no points here.
For $x=7$ we get $y^2=8$   Not in the table (not a complete square) hence no points here.
For $x=8$ we get $y^2=6$   Not in the table (not a complete square) hence no points here.
For $x=9$ we get $y^2=4$   Exists in table (hence it is a complete square) So corresponding points are: $(9,2)$ and $(9,9)$
For $x=10$ we get $y^2=0$   Exists in table (hence it is a complete square) So corresponding point is: $(10,0)$
So all the affine points are
$(0,1)$ $(0,10)$
$(1,0)$
$(2,2)$$(2,9)$
$(9,2)$ $(9,9)$
$(10,0)$
Note that we can calculate solutions for $x = 0, 1, ... , \frac{11-1}{2}$ and use symmetry in $x$ in $\mod 11$ ($10 \equiv -1 \mod 11$, $9 \equiv -2 \mod 11$, etc.) as well to obtain all points, namely $(x,y) =&gt; (x, -y) \wedge (-x, y) \wedge (-x, -y)$.
In the example above you get all points calculating all solutions until $x=5$. The solution for $x=9 \equiv -2 \mod 11$ is obtained using symmetry for $x=2$ ($(2,2) =&gt; (2, -2) = (2, 9) \wedge (-2, 2) = (9, 2) \wedge (-2, -2) = (9, 9)$) and the solution for $x = 10 \equiv -1 \mod 11$ is obtained using symmetry for $x=1$.