Formula for.G02 G03 arc centre (x,y)

I am getting rusty with the math...

So:
Can someone please give me the formulas for the x and y for a circle
based on the start x,y end x,y and radius?

I know I need to solve 2 simultaneous equations to get the answer but they are quadratic and I just can't do that anymore

math.stackexchange.com/questions/1822387...r-radians-or-degrees

Hmm, thanks but it seems that calculates the arc length not the centre point coordinates.

The equation of a circle centred on the origin is
x^2 + y^2 = r^2

For a circle not centred on the origin:

(x -a)^2 + (y - b)^2 = r^2
where (a,b) is the centre point.

We have two simultaneous equations

(Sx-Cx)^2 + (Sy-Cy)^2 = r^2
(Ex-Cx)^2 + (Ey-Cy)^2 = r^2

And by cheating, and putting this into Matlab, I got the surprisingly complicated:

Cx = (Ex^2 + Ey^2 - Sx^2 - Sy^2 - (Ey*(Ex^2*Ey + Ey*Sx^2 + Ex^2*Sy - Ey*Sy^2 - Ey^2*Sy + Sx^2*Sy + Ex*(-(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2)*(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2 - 4*r^2))^(1/2) - Sx*(-(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2)*(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2 - 4*r^2))^(1/2) + Ey^3 + Sy^3 - 2*Ex*Ey*Sx - 2*Ex*Sx*Sy))/(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2) + (Sy*(Ex^2*Ey + Ey*Sx^2 + Ex^2*Sy - Ey*Sy^2 - Ey^2*Sy + Sx^2*Sy + Ex*(-(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2)*(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2 - 4*r^2))^(1/2) - Sx*(-(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2)*(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2 - 4*r^2))^(1/2) + Ey^3 + Sy^3 - 2*Ex*Ey*Sx - 2*Ex*Sx*Sy))/(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2))/(2*Ex - 2*Sx)

Cy = (Ex^2*Ey + Ey*Sx^2 + Ex^2*Sy - Ey*Sy^2 - Ey^2*Sy + Sx^2*Sy + Ex*(-(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2)*(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2 - 4*r^2))^(1/2) - Sx*(-(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2)*(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2 - 4*r^2))^(1/2) + Ey^3 + Sy^3 - 2*Ex*Ey*Sx - 2*Ex*Sx*Sy)/(2*(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2))

For one solution and

Cx = (Ex^2 + Ey^2 - Sx^2 - Sy^2 - (Ey*(Ex^2*Ey + Ey*Sx^2 + Ex^2*Sy - Ey*Sy^2 - Ey^2*Sy + Sx^2*Sy - Ex*(-(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2)*(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2 - 4*r^2))^(1/2) + Sx*(-(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2)*(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2 - 4*r^2))^(1/2) + Ey^3 + Sy^3 - 2*Ex*Ey*Sx - 2*Ex*Sx*Sy))/(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2) + (Sy*(Ex^2*Ey + Ey*Sx^2 + Ex^2*Sy - Ey*Sy^2 - Ey^2*Sy + Sx^2*Sy - Ex*(-(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2)*(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2 - 4*r^2))^(1/2) + Sx*(-(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2)*(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2 - 4*r^2))^(1/2) + Ey^3 + Sy^3 - 2*Ex*Ey*Sx - 2*Ex*Sx*Sy))/(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2))/(2*Ex - 2*Sx)

Cy = (Ex^2*Ey + Ey*Sx^2 + Ex^2*Sy - Ey*Sy^2 - Ey^2*Sy + Sx^2*Sy - Ex*(-(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2)*(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2 - 4*r^2))^(1/2) + Sx*(-(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2)*(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2 - 4*r^2))^(1/2) + Ey^3 + Sy^3 - 2*Ex*Ey*Sx - 2*Ex*Sx*Sy)/(2*(Ex^2 - 2*Ex*Sx + Ey^2 - 2*Ey*Sy + Sx^2 + Sy^2))


For the other.

I checked numerically against a CAD drawing, the answers were correct.

Trigonometry based on the observation that the centre points lie on the perpendicular bisector of the line joining start to end points might give a simpler solution.

Can Bertd make a simple drawing of his 2 simultaneous equations ? Then we can put it into Matlab and see what happens.

Grotius wrote: Can Bertd make a simple drawing of his 2 simultaneous equations ? Then we can put it into Matlab and see what happens.

The two simultaneous equations are in my post. Solve for Cx and Cy for known Sx, Sy, Ex, Ey and r.