Let $C$ be the set of all points on the unit circle. We are given the following set \begin{equation*}A:=\left \{5v \mid v\in C\right \} \end{equation*}
$v$ is of the form $(x,y)$ for which $x^2+y^2=1$, or not?
Then $5v$ is $(5x,5y)$. Then should it satisfy the equation $(5x)^2+(5y)^2=1$ ?
Going with "on the unit circle", \begin{align*} C &= \{(x,y) \in \Bbb{R}^2 : x^2 + y^2 = 1 \} \\ A &= \{5v : v \in C \} \\ &= \{5(x,y) \in \Bbb{R}^2 : (x,y) \in C \} \\ &= \{5(x,y) \in \Bbb{R}^2 : x^2 + y^2 = 1 \} \\ &= \{(5x,5y) \in \Bbb{R}^2 : x^2 + y^2 = 1 \} \\ &= \{(5x,5y) \in \Bbb{R}^2 : \frac{(5x)^2}{5^2} + \frac{(5y)^2}{5^2} = 1 \} \\ &= \{(5x,5y) \in \Bbb{R}^2 : (5x)^2 + (5y)^2 = 25 \} \\ &= \{(\hat{x},\hat{y}) \in \Bbb{R}^2 : \hat{x}^2 + \hat{y}^2 = 5^2 \} \text{,} \\ \end{align*} which is the circle of radius $5$.