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Thread: How to find radius of circle knowing versine and length of x (diagram included)

  1. #1
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    How to find radius of circle knowing versine and length of x (diagram included)

    So we know the versine length, and we know the length of x (blue line). We also know the angle theta. From this information it is simple to find the lengths of the dotted blue/green line (idk if we need that info or not). How can I find the radius of this circle? I am pretty sure that I can find the radius if I can find L, so maybe that is a necessary intermediate step.

    Thank you in advance. You may need to click on the image to enlarge.

    How to find radius of circle knowing versine and length of x (diagram included)-versine_find_l.png
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    Re: How to find radius of circle knowing versine and length of x (diagram included)

    Let $r$ be the radius of the circle. Let $y$ be the length of the versine. Then the distance from the point where $x$ touches the circle to the center of the circle is $r$ (it is a radius). If we were to assign coordinates to that point (with the circle centered at the origin), then the point where the segment labeled $x$ intersects the segment labeled $y$ would have coordinates $(r-y,0)$. Then the point where the segment labeled $x$ intersects the circle would have coordinates $(r-y+x\cos \theta,x\sin \theta)$. This means that:

    $$(r-y+x\cos \theta)^2+(x\sin \theta)^2 = r^2$$

    Solving for $r$ gives:

    $$r = \dfrac{x^2+y^2-2xy\cos \theta}{2y-2x\cos \theta}$$

    So long as $y \neq x\cos \theta$. However, that only happens when $\theta=0$ and $x=y$.
    Thanks from topsquark, IT10101 and Vinod
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    Senior Member Vinod's Avatar
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    Re: How to find radius of circle knowing versine and length of x (diagram included)

    Quote Originally Posted by SlipEternal View Post
    Let $r$ be the radius of the circle. Let $y$ be the length of the versine. Then the distance from the point where $x$ touches the circle to the center of the circle is $r$ (it is a radius). If we were to assign coordinates to that point (with the circle centered at the origin), then the point where the segment labeled $x$ intersects the segment labeled $y$ would have coordinates $(r-y,0)$. Then the point where the segment labeled $x$ intersects the circle would have coordinates $(r-y+x\cos \theta,x\sin \theta)$. This means that:

    $$(r-y+x\cos \theta)^2+(x\sin \theta)^2 = r^2$$

    Solving for $r$ gives:

    $$r = \dfrac{x^2+y^2-2xy\cos \theta}{2y-2x\cos \theta}$$

    So long as $y \neq x\cos \theta$. However, that only happens when $\theta=0$ and $x=y$.
    Hello,
    thanks
    Last edited by Vinod; Aug 3rd 2018 at 12:00 AM.
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