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I have circle which I know intersects the x axis at -11.5 and 11.5. It intersects the Y axis at 1. How can I calculate the (positive) Y value for any X value between -11.5 and 11.5?

This is to calculate an arc for a carpentry project. the arc is too shallow to use the usual mechanical techniques for finding the radius.

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Seen this? –  J. M. May 6 '13 at 17:36
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1 Answer

Since the intersection with the $x$ axis are symmetrical with respect to $0$, it is easy to prove that the centre lies on the $y$ axis.

Let $(0,a)$ be the centre. Then the equation is

$$x^2+(y-a)^2=R^2 (*)\,.$$

$$11.5^2+a^2=R^2=(1-a)^2 \,.$$

This Yields:

$$11.5^2=1-2a$$

You can find $a$ from here, get $R$ from $R^2=11.5^2+a^2$ and then get your equation $(*)$.

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or simply $R = 1+a$ –  karakfa May 6 '13 at 18:10
    
This worked perfectly. Thank you very much. –  Peter de Klerk May 6 '13 at 18:36
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