Pretty complicated expressions for h and k. This can all be done symbolically, of course, but you'll get some Finally, set:Īnd you'll have everything you need to know about the circle. Solve these, and you'll have the coordinates Two equations, leaving you with two simultaneous linear equations in That will eliminate r, h^2, and k^2 from the last To solve these, subtract the firstįrom the other two. Since the three points all lie on the circle, their coordinates will Find the circumference of this circle: r 10 units 2 5 units. We then use the appropriate value in this equation: C 2 r (where r represents radius, of course). Let (h,k) be the coordinates of the center of the circle, and r its In order to calculate a circle’s circumference, we need to know either its diameter or its radius. Subject: Re: finding the coordinates of the center of a circle (x1,y1), (x2,y2) and (x3,圓) - how do I find the coordinates of theĬenter of a circle on whose circumference the points lie? Subject: finding the coordinates of the center of a circleĬan you help me? If I have the x and y coordinates of 3 points - i.e. Finding the Center of a Circle Given 3 Points Since you have three points which are exactly on the arc former by the circle centered at (0,0) (given), the system can be solved exactly rather than requiring a least-squares approximation. The solution to this is almost identical to the "circle of best fit for a non-over-determined system". It will tell you if B lies on the left or the right of AC. Hint: consider the sign of the area of the triangle ABC, precisely the half of the determinant of the system. The radius and angles can be found using the Cartesian-to-polar transform around the center: R= Sqrt((Xa-X)^2+(Ya-Y)^2)īut you still miss one thing: what is the relevant part of the arc ? Smaller or larger than a half turn ? From Ta to Tb or from Tb to 2 Pi to Ta + 2 Pi, or what ? The answer is much less obvious than it seems, try it (because the three angles Ta, Tb and Tc are undeterminate to a multiple of 2 Pi and you cannot sort them) ! This linear system of two equations in two unknowns is easy to solve with Cramer's rule.
Subtracting the first member from the second and the third, we get after regrouping: 2(Xa-Xb) X + 2(Ya-Yb) Y + Xb^2+Yb^2-Xa^2-Ya^2 = 0Ģ(Xa-Xc) X + 2(Ya-Yc) Y + Xc^2+Yc^2-Xa^2-Ya^2 = 0
The center of the circle is equidistant to the three given points: (X-Xa)^2+(Y-Ya)^2 = (X-Xb)^2+(Y-Yb)^2 = (X-Xc)^2+(Y-Yc)^2
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