# Determine center point of offset circle

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1. Oct 5, 2014

### examorph

I am working on an algorithm which requires the coordinates for the center point of an offset circle. Dimensions available to find this are shown in the image below and the dimension required is labeled as X:

The point at the very left of the arc is a quadrant therefore, the circle center also 2.27 away from the line with a dimension of 3. The actual value of x in this example is 2.47 (rounded), this was found using CAD however, I am trying to find an mathematical relation for this.

2. Oct 5, 2014

### Mentallic

Sorry, what is the significance of the line with length 3? At the moment all I'm seeing is an arc of a partial circle and then a line which doesn't seem to have any bearing on the characteristics of the circle.

3. Oct 5, 2014

### examorph

I apologize for the confusion. Where I said "Circle" I should have said "Arc". All dimensions shown above are just what are available and not necessarily needed to determine the center point of the arc however, I think that in the case of the line with length 3, it would be useful as the combination of this with the distance 2.27 shows coordinates to a point on the arc.

4. Oct 5, 2014

### Mentallic

It's an arc of a circle just to be clear though. We could always have drawn in the rest of this circle with dotted lines. So then the right end of the line segment of length 3 lies on this circle, correct?

edit: But then if that were the case, then the value of x would simply be (4.5-3)+3/2 = 3.

Unless there is some particular reason as to why the segment has a given length of 3, then the question can't be answered.

Last edited: Oct 5, 2014
5. Oct 6, 2014

### TheoMcCloskey

Actaully, there is sufficient information. Recall the 4.5 dimension.

OP indicates that radius is tangent at upper left point (let's call it point "A").
Let's use that point (A) as an origin in the x,y plane.

The equation of a circle is (x-a)^2 + (y-b)^2 = r^2 where point (a,b) is the center of the circle of radius r.

Since circle is tangent at point A, then a = r and the center lies on this axis, thus, b = 0.

The coorindates of the lower right point (let's call it "B") is as follows
x_b = 4.5 - 3.0 = 1.5
y_b = -2.27

Thus, the circle must satisfy
(x_b - a)^2 + (y_b - b)^2 = r^2
or
(x_b - r)^2 + (y_b - 0)^2 = r^2

==> r = ( (x_b)^2 + (y_b)^2 ) / ( 2 * x_b)

numerically,
r = 1.5^2 + (-2.27)^2 / (2 * 1.5)
= 2.47

6. Oct 9, 2014

### Mentallic

Ahh you're right! Good work!