At time $t=0$ a particle in uniform circular motion in the plane has velocity $\vec{v}=6\vec{i}-4\vec{j}$ and acceleration $\vec{a}=2\vec{i}+3\vec{j}$. Find the radius and center of its orbit if at time $t=0$ it is at the point $P=(0,0)$.
The radius of its orbit is found using the formula, $R=\frac{\left\|\vec{v}\right\| ^{2}}{\left\|\vec{a}\right\|}$.
Thus, $R=\frac{52}{\sqrt{13}}$.
How to find the center of its orbit?
Hint:
The centre will lie along the acceleration vector.Thus,centre(C) will be $(R\cos A,R\sin A )$,where A is the angle which PC makes with positive $x$-axis.