Homogeneous polynomials' roots I'm trying to understand the proof of this result
Let $f(x,y) \in K[x,y]$ be an homogeneous polynomial s.t. $deg(f) = d>0$ then $\exists$ at most d $(a,b) \in K^2$ non-trivial roots of $f$
Proof:

$f(x,y)=\sum_{i=0}^{d} c_ix^iy^{d-i}$ 
$deg(f)>0 \implies \exists c_i \neq 0 \:$. Since f is homogeneous, $f(a,b)=0 \implies f(ta,tb)=0 \ \forall t \in K$
B.C. $(1,0)$ is not a root $\implies c_d \neq 0$. Then $f(t,1)$ is an homogeneous polynomial in one variable, with $deg=d$ $\implies$ it as at most $d$ roots
$f(t,1)=a_0\prod_{i=1}^{d}(t-a_i)$. Let's say $\ t = \frac{x}{y}$, then 
$f(t,1)=f(\frac{x}{y},1)=a_0\prod_{i=1}^{d}(\frac{x-a_i}{y})$ but $f(\frac{x}{y},1)=y^df(x,y)$, which has a root in $(1,0)$ (contradiction)
So $(1,0)$ must be a root. Let $r<d$ be its multiplicity, then
$c_d = c_{d-1} = ... = c_{d-r+1}=0$
$f(x,y)=\sum_{i=0}^{d-r} c_ix^iy^{d-i}=y^{r}\sum_{i=0}^{d-r}c_ix^iy^{d-r-i}$
And that's how it ends.
I can't understand how this works:
1)I don't think it's necessary that $f(x,y)$ has a root in $(1,0)\ $ (for example $f(x,y)=x^d+y^d$ doesn't have it), so why he has followed this line? Also, he substitutes $t$ with $\frac{x}{y}$ but weren't we considering $f(t,0)$? Is this legit?
2)How does this proves the thesis?
Thank you for help
 A: Thanks to RandyMarsh, think I can close the question now, if the following should be correct. Here is the adjusted statement
Let $f(x,y)\in K[x,y]$ be an homogeneous polynomial s.t. $deg(f) = d>0$, then $\exists$ at most $d\ (a,b) \in K^2\ projective\ non-trivial \ roots\ [(a,b) \neq (0,0)]$ of $f$, counted with their multiplicity.
Proof:
$f(x,y)= \sum_{i=0}^{d} c_ix^iy^{d-i}\\deg(f)=d>0 \implies \exists c_i \neq 0\\ \text{Since } f \text{ is homogeneous, } f(a,b)=0 \implies f(ta,tb)=0\  \forall t \in K\\ \text{We have two cases} \\ 
\text{1) }(1,0)\text{ is not a root}\implies c_d \neq 0 \text{. Then } f(t,1) \text{ is an one-variable polynomial of } deg = d \implies \text{ it has at most d roots}\\
2)(1,0) \text{ is a root. Let } r \leq d \text{ be its multiplicity, then } c_d = c_{d-1} = \dots = c_{d-r+1} = 0 \\
\implies f(x,y) = y^r \cdot g(x,y) \text{ where g is homogeneous with } deg(g)=d-r \implies g(1,y) \text{ is an one-variable polynomial of } deg=d-r \text{, so it has at most } d-r \text{ roots.}$.
I considered two distinct cases because i didn't see the point assuming that $(1,0)$ must be a root.
