Find the radius of a circle from a chord

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To find the radius of a circle using the length of a chord (c) and the distance from the midpoint of the chord to the circle's edge (m), the equation is derived from right triangle relationships. The equation can be expressed as radius = (c^2/4 + m^2)/(2m). By drawing a right triangle with the chord and the line from the midpoint to the circle's edge, trigonometric identities can be applied. Specifically, using sin(theta) and cos(theta) helps establish the relationship needed to derive the radius. This method effectively combines geometry and trigonometry to solve for the radius.
Lizardjuice7
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Hi,

Can someone please explain to me how you get the equation:

radius=(\sqrt{c}+m2)/2m

c=length of chord

m=distance from midpoint of chord to edge of the circle

I would like to find the radius of the circle from only knowing those two quantities.

I have found the equation on various websites across the web (including http://www.math.utah.edu/~eyre/rsbfaq/physics.html), but I do not know how it was derived.

Thanks,

Lizardjuice7
 
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From the link you gave they say the equation is r = ((c^2)/4 + m^2)/(2m), which is very good because that's the answer i just got from solving the problem. (EDIT: I think I see the problem, the link gives c^2/4 without parenthesis so you probably assumed c^(2/4) instead of (c^2)/4)

Draw your circle with a chord and a line from the midpoint of the chord to the opposite side of the circle. Chord is of length c and line bisecting the chord is of length m.

Now draw a line connecting the midpoint of the circle (which is on line m) to one of the corners of the chord. Because line m bisects the chord, this is a right triangle.

Consider the angle theta between the chord and the origin (the angle on your triangle which does not on the origin). Using normal trig relations:

sin(theta) = (m - r)/r
cos(theta) = c/(2*r)

Now just compute sin(theta)^2 + cos(theta)^2 = 1 and do the algebra!
 
thanks, that's a big help
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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