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How does JSL calculate whether the fourth point is on the same circle as the first three points?

lwx228

See a method of Craige_Hales: Smallest Covering Circle
But I still haven't learned.
My need is to determine a circle by the first three points.
And then determine if the fourth point is on the circle.

So how to write JSL?

Thanks Experts!

Accepted answers

ian_jmp

@Craige_Hales solution at your link above does all the hard work, and it calculates 'center' and 'radius' for the circle. So, if you want to test a specific point, you could add something like this to his code:

// Test if a specific point is on the circle (in this case, the origin)
myPoint = Transpose([0, 0]);
myPointShifted = myPoint - center;
myDistance = Sqrt(myPointShifted[1]^2 + myPointShifted[2]^2);
tol = 1E-6;
if(abs(myDistance - radius) <= tol,
	Print("Point is on the circle"),
	Print("Point is not on the circle")
	);

Craige_Hales

See if Circle through three points is the answer.

All comments

ian_jmp

// Test if a specific point is on the circle (in this case, the origin)
myPoint = Transpose([0, 0]);
myPointShifted = myPoint - center;
myDistance = Sqrt(myPointShifted[1]^2 + myPointShifted[2]^2);
tol = 1E-6;
if(abs(myDistance - radius) <= tol,
	Print("Point is on the circle"),
	Print("Point is not on the circle")
	);

lwx228

Yes, thanks to Craige_Hales for doing so much work.
Thanks for ian help!

I tried to modify the code to change the number of points generated in line 1 to three.
I found that this code only drew the circle with the largest diameter.

I really don't know how to modify it so that the circle is determined by A, B and C at the same time.
So I still need help.

Thanks Experts!

Craige_Hales

See if Circle through three points is the answer.

lala

That's a good question.
I tried the new method provided by @Craige_Hales ( Circle through three points ) , combined with the ian_jmp'code. But it doesn't get the job done.

Can @ian_jmp modify the following code according to the new code of Craige_Hales?Thank you very much!

ian_jmp

As mentioned in my post, the first post from @Craige_Hales calculates the variables 'center' and 'radius' and my code snippet above uses these.

His new code contains these lines (near the very end):

// choose distance to A, could use B or C too...
Circle( {xc, yc}, Sqrt( (xc - ax) ^ 2 + (yc - ay) ^ 2 ) );

Looking in 'Help > Scripting Guide' for 'Circle', you will see that '{xc, yc}' are the coordinates of the center and the result of the 'Sqrt()' operation is the radius (of the circle that's currently drawn).

lala

OK、Thanks!

Circle( {20, 20}, 4, 7, 10/* no fill for concentric circles */ );