Suppose we are given in $R^2$ an angle ACB and a circle centered at C, like the following picture shows.
How can we "draw" a $C^1$ curve containing the segments outside of the circle and the part AB would be inside the circle and angle at the same time?


You can easily solve for a quadratic polynomial that will make this $C^1$ (or, indeed, a quartic polynomial that will make it $C^2$). Just take $y=f(x)=ax^2+bx+c$, set $f(\pm x_0)=0$ and $f'(\pm x_0) = \mp m$. You have to specify $x_0$ and $m$ appropriately to make sure the curve stays below $C$. (Here I'm taking the $x$-coordinate of $C$ to be $0$ and the $x$-coordinates of $A$ and $B$ to be $\pm x_0$ respectively.)