MHB Finding the Center of a Circle

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To find the center of a circle with a known diameter and a single point on the circumference, at least two distinct points on the arc are required. By drawing a straight line between these two points and then bisecting that line, you can create a perpendicular bisector. Repeating this process with another pair of points will allow the intersection of the bisectors to determine the circle's center. A simple formula cannot be provided without additional points, as one point alone is insufficient for accurate calculation. Understanding these geometric principles is essential for locating the center effectively.
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Hello!

I have an application where I need to find the center of a circle where I am having trouble coming up with a simple way to do this. The diameter of the circle is known and i want to be able to determine the location of it where only a portion of the circle is known. (see the image below) I will know the location of a single point anywhere in the red circumference could anyone help me with this?
1623225283571.png
 
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Draw the straight line between two points on the circle. Draw the line bisecting that straight line. Do the same with a second pair of points. The two bisectors intersect at the center of the circle.
 
Thank you for your reply! I am looking for a formula that i could use to calculate where the center of the circle is based on the known diameter and a single point that resides in the red area. Sorry i should have meant to say in the original post that i was looking for a simple formula.
 
Runner74 said:
Thank you for your reply! I am looking for a formula that i could use to calculate where the center of the circle is based on the known diameter and a single point that resides in the red area. Sorry i should have meant to say in the original post that i was looking for a simple formula.

You're going to need at least two distinct points on the red arc of the circle to find its center.
 

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Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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