How to find the equation of $X$? A particle Q is moving at a constant speed $V$ in a circular path of radius $R$. P is a fixed point below $O$(the center) at a distance $r$.
X$=$PQ
How can I find an equation for $X$?
It is actually a part of a dynamics(motion of a particle in 2D) problem & I can't think of a way to find a mathematical relationship between $R$ and $r$.
I forgot to tell about $\theta$, the angle $PQ$ makes with the horizontal line.
 A: Use the cosine theorem on triangle POQ:
$x^2=r^{2}+R^{2}-2Rr\cos\left(\frac{v}{R}t+\frac{\pi}{2}\right)$
A: Consider a coordinate system with $O$, the center of the circle, as the origin. The radius of the circle is $R$, and the speed of the particle $Q$ is the constant $v$. Assuming the particle is at $(R, 0)$ at time $t = 0$, and then moves along the circle in the counter-clockwise direction:
$\vec{Q} = R \cos\left( \dfrac{v}{R} t \right) \hat{i} + R \sin \left( \dfrac{v}{R} t \right) \hat{j}$
Now, $P$ is directly below $O$ at a distance $r$ from it, so $P = -r\hat{j}$. Thus
$\begin{align}
\vec{PQ} & = \vec{Q} - \vec{P}\\
& = R \cos\left( \dfrac{v}{R} t \right) \hat{i} + \left[ R \sin \left( \dfrac{v}{R} t \right) + r \right] \hat{j}
\end{align}$
The distance $PQ$ is $|\vec{PQ}|$, so
$\begin{align}
PQ & = \sqrt{R^2 \cos^2 \left( \dfrac{v}{R} t \right) + \left[ R \sin \left( \dfrac{v}{R} t \right) + r \right]^2 }\\
& = \sqrt{ R^2 \cos^2 \left( \dfrac{v}{R} t \right) + R^2 \sin^2 \left( \dfrac{v}{R} t \right) + 2rR \sin \left( \dfrac{v}{R} t \right) + r^2}
\end{align}$
$\boxed{PQ = \sqrt{R^2 + r^2 + 2rR \sin \left( \dfrac{v}{R} t \right)}}$
A: I may be wrong but if you need x as a distance , can't we use Pythagore theorem .
In that case the equation would be 
$$
x=\sqrt(PQ^2)=
\sqrt(R^2+r^2)
$$
But that equation does not involve the speed constant.
