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 $D(200,Y)$.


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

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


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




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:


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


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



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