Is it possible to have an equation in the form of $$ax^2 + by^2 = q^2 +cx$$ or $$f(x)=\sqrt{\frac{q^2+cx-ax^2}{b}}$$
such that
$$f(k)=0; f(m)=0$$
where the distance between $(m,0)$ & $(k,0)$ is 1 and where $m,k>1$ and the area under the curve above the x-axis is $\frac{\pi}{2}$?
In other words, I want to know if it's possible to move a function of a circle with radius $r=1$ over the x-axis