How to make a perfectly circular turn in less than 90 degrees

(The examples are for a 45 degree turn, try it for yourself! )
1.Make an isosceles triangle with 3 vertices with right angle trigonometry
Example points(x,z): (-10,0)(0,0) (7.07107, 7.07107)

2.Plug in the angle of turn (a 45 degree turn makes up an 8th of a circle) into this equation:
Cos(angle of turn/2)
Example: cos(45/2)=0.9238795325

3.Enter this value as the weight for the the vertex in the middle of the turn
Example: In the vertex (0,0), enter 0.9238795325 into the weight slot (default is 1)

For a visual, turn on the Radius Comb Single
This is showcased in a roller coaster I've uploaded called "The Dips."

Let us also not forget the left and right turn elements that come with the game

^ That's a good point, but these aren't editable after being placed which for something as basic as a circular arc seems like a handy option to have.

Btw, if someone is interested, a derivation of the formula in question can be found here. In that case it's sin(angle/2) instead of cosine because it's using the angle inbetween two edges of the control polygon. Oh, and that formula works for every angle inbetween 0 and 180°.

visualized:

Thanks bestdani, that looks amazing!