Trigonometry sum of solutions question 
Problem: For which $a$ will the sum of solutions be equal to $100$, in $\sin(\sqrt{ax-x^2})=0$.

The attempt at a solution:
For $\sin(x)=0$, $x$ must be equal to $0$, so we get $$\sqrt{ax-x^2}=0\\ax-x^2=0\\x(a-x)=0\\x=0\\a-x=0$$last 2 are the solutions as I understand but I can't seem to finalize the solution, please help. Thank you in advance.

I think there is no such $a$. Note that
$$\sin(\sqrt{ax-x^2}) = 0 \iff \sqrt{ax-x^2}=k \pi$$
for some $k \in \mathbb{N}$. Squaring both sides of the second equation and solving the second degree equation shows
$$x_{1,k} = \frac{a}{2} +\frac{\sqrt{a^2-4k^2\pi^2}}{2} \text{ and }x_{2,k} = \frac{a}{2} -\frac{\sqrt{a^2-4k^2\pi^2}}{2}$$
In particular we have $x_{1,k}+x_{2,k} = a$. These solutions exists as long as 
$$a^2 - 4\pi^2k^2 \geq 0 \iff \frac{|a|}{2\pi} \geq |k|.$$
In particular if $N$ denotes the integer part of $\frac{|a|}{2\pi}$, then the sum of all solutions is give by $S=a2N$. Asking for $S=100$ leads to the equation
$$a(2\lfloor |a|/2\pi\rfloor+1) = 100,$$
where $\lfloor \cdot \rfloor$ denotes the floor function. It follows that there is no solution.
