Finding the Center of a Circle from 2 PointsDate: 06/01/99 at 08:24:55 From: John Subject: Need to find centre of circle from X, Y, and r If you have 2 points on a circle, both X and Y figures are known, and a radius is given, how is possible to find the centre of the circle? Date: 06/01/99 at 10:07:40 From: Doctor Rob Subject: Re: Need to find centre of circle from X, Y, and r Using the two given points as centres, draw two circles, each with the given radius. The places where those two circles intersect are the possible centres of the original circle. Algebraically, if the two points are (a,b) and (A,B), and the radius is r, you can solve the two equations (x-a)^2 + (y-b)^2 = r^2, (x-A)^2 + (y-B)^2 = r^2, simultaneously. (Hint: Subtract one from the other, and you'll have a linear equation as a result. Solve that for one of the variables, and substitute in either of the above two equations. That will give you a quadratic equation in the other variable.) This works because the distance from the centre of the circle to each of (a,b) and (A,B) is r. Example: Find the center of the circle passing through (1,4) and (5,1) with radius 13/2. (x-1)^2 + (y-4)^2 = (13/2)^2, (x-5)^2 + (y-1)^2 = (13/2)^2, expanding these we get x^2 - 2*x + 1 + y^2 - 8*y + 16 = (13/2)^2, (-) x^2 - 10*x + 25 + y^2 - 2*y + 1 = (13/2)^2, -------------------------------------------- subtracting these 8*x - 24 - 6*y + 15 = 0, y = (8*x-9)/6, substituting this (x-5)^2 + ([8*x-15]/6)^2 = (13/2)^2, x^2 - 10*x + 25 + (16/9)*x^2 - (20/3)*x + 25/4 = 169/4, (25/9)*x^2 - (50/3)*x - 11 = 0, 25*x^2 - 150*x - 99 = 0, (5*x - 33)*(5*x + 3) = 0, so x = 33/5 or x = -3/5, y = 73/10 or y = -23/10. Thus the center of the circle could be at either (33/5,73/10) or (-3/5,-23/10). You can check that these points are 13/2 units away from both (1,4) and (5,1). - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ Date: 06/15/2014 at 06:17:10 From: Josh Subject: find all possible circles given 2 points (but no radius) After reading this, it occurs to me that the circle with minimum radius is the midpoint between the two circles. And I *think* that -- given any other radius greater than this min -- the two alternative points lie on the line that is perpendicular to the line connecting the two points at the midpoint. If that's true, it might be a useful diagram for people to visualize the general solution. Just a suggestion ... Thank you for the site in any case! - J. |
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