Radius in a u.c.m.

Data

Center of the circular trajectory: C (0,0) m

Point belonging to the circular trajectory: A (3,4) m

Resolution

a) The position vector $\overrightarrow{r}$ of a body is a vector that goes from the origin of coordinates to the position of said body. So, to calculate the vector we must calculate the vector that goes from the Center C to point A.

$$\begin{gathered} \overrightarrow{r}=\overrightarrow{CA}=\left(3,4\right) - \left(0,0\right)=\left(3,4\right)\Rightarrow \\ \boxed{\overrightarrow{r}=3 ·\overrightarrow{ i} + 4 ·\overrightarrow{ j }m} \end{gathered}$$

b) To calculate the radio of the circumference we must calculate the distance from the origin C to any of the points that make up the trajectory. Since we know one of those points (A) and we already have the position vector for said point, the magnitude of said vector is equal to the value of R. Therefore:

$$\begin{gathered} \left|\overrightarrow{r}\right|=R=\sqrt{3^2+4^2}=\sqrt{25} \Rightarrow \\ \boxed{R=5 m} \end{gathered}$$

c) Considering that we know the position vector of point A (3.4):

We have two equations to calculate the angular position:

$$\left.\begin{array}{r}x =\hspace{0.17em}R· \cos \phi \\ y = R · \sin \phi \end{array}\right\}\left.\begin{array}{r}3=5·\cos \phi \\ 4=5·\sin \phi \end{array}\right\}$$

Using the first one:

$$\begin{gathered} \cos \phi =\hspace{0.17em}\frac{3}{5} \Rightarrow \\ \phi = arc \cos \frac{3}{5} \Rightarrow \\ \boxed{\phi = 53.13º} \end{gathered}$$