# Circle circumference point calculation

In the image below, I have a part of a circle. Given,

$$\text{chord }d=1050\ mm\\ \text{height }f=50\ mm\\ \text{radius }R=2781\ mm\\ \text{centre }O(700\ mm,2781\ mm)$$

and $$3$$ points $$A(0,200), B(1050,200), C(525,150)$$.

I would like to know if there is an equation to calculate the $$y$$ coordinate for any given $$x$$ coordinate, for example the point $$D(200,Y)$$.

• I would start by figuring out the equation of the circle. I mean, in the form $$(x-x_0)^2 + (y-y_0)^2 = R^2$$ – Matti P. Jan 14 '19 at 13:20
• The question has already been asked here. – whiskeyo Jan 14 '19 at 13:22

Try to solve something more general: Suppose that you have 3 arbitrary points $$A(x_A,y_A)$$, $$B(x_B,y_B)$$ and $$C(x_C,y_C)$$. Let's find center $$O(x_O,y_O)$$ and radius $$R$$ of the circle passing through all three points.

Denote midpoints of segments $$AB$$,$$BC$$ with $$P$$,$$Q$$:

$$P(x_P=\frac{x_A+x_B}{2}, \ y_P=\frac{y_A+x_B}{2})$$

$$Q(x_Q=\frac{x_B+x_C}{2}, \ y_Q=\frac{y_B+y_C}{2})$$

Obviously $$OP\bot AB$$ and $$OQ\bot BC$$ which means that:

$$(x_O-x_P)(x_B-x_A)+(y_O-y_P)(y_B-y_A)=0$$ $$(x_O-x_Q)(x_C-x_B)+(y_O-y_Q)(y_C-y_B)=0$$

or:

$$(x_O-\frac{x_A+x_B}{2})(x_B-x_A)+(y_O-\frac{y_A+y_B}{2})(y_B-y_A)=0$$ $$(x_O-\frac{x_B+x_C}{2})(x_C-x_B)+(y_O-\frac{y_B+y_C}{2})(y_C-y_B)=0$$

or:

$$(x_B-x_A)x_O+(y_B-y_A)y_O=\frac{x_B^2-x_A^2+y_B^2-y_A^2}{2}\tag{1}$$

$$(x_C-x_B)x_O+(y_C-y_B)y_O=\frac{x_C^2-x_B^2+y_C^2-y_B^2}{2}\tag{2}$$

The point is: equations (1) and (2) are linear wtih two unknowns $$(x_O,y_O)$$ and, assuming that points $$A,B,C$$ are not collinear, you can always find a unique solution. You can also come up with a general formula for $$(x_O,y_O)$$ but I leave it up to you as an exercise.

Once you have $$x_O,y_O$$ you can find the radius of the circle:

$$R=\sqrt{(x_A-x_O)^2+(y_A-y_O)^2}$$

For any point on the circle $$K(x_K,y_K)$$:

$$(x_K-x_O)^2+(y_K-y_O)^2=R^2$$

And if you know $$x_K$$, you can calculate the value of $$y_K$$:

$$y_K=y_O\pm\sqrt{R^2-(x_K-x_O)^2}$$