Maximum cycle in a graph with a path of length $k$ I don't understand why this stands:

Let $G$ be a graph containing a cycle $C$, and assume that $G$ contains a path of length at least $k$ between two vertices of $C$. 
  Then $G$ contains a cycle of length at least $\sqrt{k}$.

Since we can extend the cycle $C$ with the vertices of the path, why don't we get a cycle of length $k+2$? ($2$ being the minimum number of vertices belonging to $C$ between the vertices where $C$ connect to it).
I really don't see where that square root is coming from.
For reference this is exercise $3$ from Chapter $1$ of the Diestel book.

Let $\{v_1,…,v_n\}$ be common vertices along the $xy$-path $P$ and the cycle $C$. Suppose $n≥\sqrt{k}$.  As a result, $C$ has at least $\sqrt{k}$ vertices, so it is a cycle of length at least $\sqrt{k}$.
However, suppose $n<\sqrt{k}$. The cycle interrupts $P$ at fewer than $\sqrt{k}$ vertices. Since $P$ has a total of $k$ edges, the length $l$ of the longest uninterrupted portion of $P$ is
$l>\frac{k}{\sqrt{k}}=\sqrt{k}$ 
Therefore, for some $v_jv_i∈\{v_1,…,v_n\}$, where $j=i+1$, there is a $v_i$$v_j$-path $P_1∈P$ that as at least $\sqrt{k}$ edges. There is also a $v_i v_j$-path $P_2∈C$ of length at least $1$, so $P_2∪P_1$ makes a cycle of at least length $\sqrt{k}$.
