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

1. ## 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.

2. ## 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 $x=y$.

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

Originally Posted by SlipEternal
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