Question: > 
> Show that the locus of the midpoint of a family of parallel chords of a circle is a diameter which is perpendicular to the given family of chords.
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Please help me understand the question with a figure .

Answer: 
The previous sentence states that if you take the numbers $a, 2a, \ldots, (p-1)a$ and write down their remainders when dividing by $p$, you will get the numbers $1,2,\ldots,p-1$ in some order.

In particular, this implies one of the original numbers $a, 2a, \ldots, (p-1)a$ is equivalent to $1$ modulo $p$. This answers your first question.

So, we have $$ab \equiv 1 \mod p.$$
If $b=1$ and $a \in S$, is this possible?
If $b = p-1$ and $a \in S$, is this possible?
Let $a,r\in \{0,1,2,\cdots ,p-1\}$. We want to argue that there exists some unique $b\in \{0,1,2,\cdots ,p-1\}$ such that $$ab\equiv r\mod p$$Assume by contrary that there exist $b\_1$ and $b\_2> b\_1$ such that $$ab\_1\equiv r\mod p\\ab\_2\equiv r\mod p$$therefore $$ab\_1\equiv ab\_2\mod p$$since $\gcd(a,p)=1$ we obtain $$b\_1\equiv b\_2\mod p$$which yields to the impossible $p|b\_2-b\_1$ since $b\_2-b\_1\in\{0,1,2,\cdots ,p-1\}$ and therefore $\gcd(p,b\_2-b\_1)=1$. This completes the proof of uniqueness of such $b$. **Now simply, let $r=1$.**