Definition of RP/poly is not clear! In Motwani and Raghavan's textbook, p38-39, they states:

A sequence $C_1, C_2, C_3, \dots$ of circuits is a circuit family for function $f_n\colon {0,1}^n \rightarrow \{0,1\}$ if $C_n$ has $n$ inputs and computes $f_n(x_1, x_2, \dots, x_n)$ at its output for all n-bit inputs $(x_1, x_2, \dots, x_n)$. The family $C$ is said to be polynomial-sized circuits if the size of $C_n$ is bounded above by $p(n)$ for every $n$, where $p(\cdot)$ is polynomial.

It is clear that this is the definition of P/poly. And the next paragraph is the definition of RP/poly.

A randomized circuit family for $f$ is a circuit family for $f$ that, in addition to the $n$ inputs $x_1, x_2, \dots, x_n$, takes $m$ random bits $r_1, r_2, \dots, r_m$, each of which is equiprobably 0 or 1. In addition, for every $n$, the circuit $C_n$ must satisfy two properties:

If $f_n(x_1, x_2, ..., x_n)=0$, then the output of the circuit is 0 regardless of the values of random inputs $r_1, r_2, \dots, r_m$.
If $f_n(x_1, x_2, ..., x_n)=1$, the the output of the circuit is 1 with probability at least $\frac12$. In other words, at least one half of the $2^m$ choices of the bits $r_1, \dots, r_m$ will result in the circuit evaluating to 1. We will refer to such $m$-tuples $r_1, r_2, \dots, r_m$ as witnesses for $(x_1, x_2, \dots, x_n)$, in that they testify to the correct value of $f_n(x_1, x_2, \dots, x_n)$ when it is 1.


Question: I don't understand what is relationship between $n$ and $m$ or $r_1, \dots, r_m$ and $f_n(x_1, x_2, \dots, x_n)$? For example, we have n inputs, i.e. $x_1, x_2, \dots, x_n$. But, in second property, they states that at least one half of the $2^m$ choices of the bits $r_1, \dots, r_m$ will result in the circuit evaluating to 1. Now, how did we bring $r_1, \dots, r_m$ to $f_n(x_1, x_2, \dots, x_n)$?
The circuit $C_n$ has $n+m$ inputs, $x_1,\ldots,x_n,r_1,\ldots,r_m$. Its size is polynomial in $n$, so without loss of generality, $m$ is polynomial in $n$. What we want is:

If $f_n(x) = 0$ then $\Pr_r[C_n(x,r) = 0] = 1$.
If $f_n(x) = 1$ then $\Pr_r[C_n(x,r) = 1] \ge 1/2$.

Here $x$ is shortcut for $x_1,\ldots,x_n$, and $r$ is shortcut for $r_1,\ldots,r_m$.