Recognize that the arc length is the same as the angle between line $\displaystyle r$ and line $\displaystyle x$.
Lines $\displaystyle r$, $\displaystyle x$ and $\displaystyle y$ form a right triangle. Thus you can use the formula:
$\displaystyle \tan(\theta)=\frac{y}{x}$
Does the question ask for the arc-length from $\displaystyle (1,0)$ to $\displaystyle (x,y)$ in the first quadrant.
If so the answer is $\displaystyle \arccos \left( {\frac{x}{{\sqrt {x^2 + y^2 } }}} \right)$.
That is the measure of the central angle times the radius which in this case is $\displaystyle R=1$.
If it could be that $\displaystyle (x,y)\in II$ the same formula works.
Now if $\displaystyle (x,y)\in III\text{ or }IV$ then we need to know if we are still measuring in a counter-clockwise direction or not.
This directive the use of calculators and trigonometric tables is not allowed is impossible to follow except for a very, very few values of $\displaystyle x~\&~y$.