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I have prove the theorem: There is only one circle passing through three given non-collinear points in both geometrical and algebraic ways. THere is one question that I just have no idea with. 'the accuracy and limitations of this technique"

I dont know what to write...! thank you

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In the geometrical way, you can use the fact that in a circle, if you draw a line between any two points in the circle and you bisect that line, the bisected line will go through the center of the circle. So if you have three points, you can draw two bisects and they will intersect at a unique point, which will be the center of your circle. To determine its radius, just compute the distance from your center to any of your three points.

In the algebraic way, you have three points $(x_1, y_1), (x_2, y_2), (x_3,y_3)$ such that $$ (x_i - a)^2 + (y_i - b)^2 = c^2, \qquad i=1,2,3 $$ which gives you three equations in three unknowns, namely $a,b,c$ (the coordinates of the center $(a,b)$ and the radius $c$). You can use one of the equations to reduce the other two to linear equations and solve the linear system to get the coordinates $a$ and $b$. Once you have those two, any of the three equations will give you $c$.

Hope that helps,

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  • $\begingroup$ we need to non-co-linearity to prove the uniqueness of the solution(mathworld.wolfram.com/LinearSystemofEquations.html) $\endgroup$ – lab bhattacharjee May 25 '13 at 13:17
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    $\begingroup$ Better yet, think of this as a system of linear equations $$x_i^2 + Ax_i + y_i^2 + By_i + C = 0\,, \quad i=1,2,3\,,$$ and check that the coefficient matrix is nonsingular if and only if the three points are noncollinear. $\endgroup$ – Ted Shifrin May 25 '13 at 13:20
  • $\begingroup$ @lab bhattacharjee : The non-colinearity will appear when one tries to solve the linear system. I was assuming OP needed only a hint. I gave him one. $\endgroup$ – Patrick Da Silva May 25 '13 at 13:23
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    $\begingroup$ There are many ways to solve linear systems. At some point you just have to do it. $\endgroup$ – Patrick Da Silva May 25 '13 at 13:23
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Geometrically, you can prove it this way:

Since the three points are noncollinear, you can construct perpendicular bisectors through each of the line segments formed between the points. And since the perpendicular bisectors all intersect at one point, that means those points are all equidistant from that intersection point. Therefore there is a circle.

Hope that helped!

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In the algebraic way, you have three points $(x_1,y_1),(x_2,y_2),(x_3,y_3)(x_1,y_1),(x_2,y_2),(x_3,y_3)$ such that
$(x_i−a)^2+(y_i−b)^2=c_2,i=1,2,3$
$(x_i−a)^2+(y_i−b)^2=c_2,i=1,2,3$
which gives you three equations in three unknowns, namely $a,b,c$ (the coordinates of the center $(a,b)(a,b)$ and the radius $c$). You can use one of the equations to reduce the other two to linear equations and solve the linear system to get the coordinates $(a,a)$ and $(b,b)$. Once you have those two, any of the three equations will give you cc.

Hope that helps,

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  • $\begingroup$ In the algebraic way, you have three points ... This, and the rest of your "answer" is just a copy/paste from @PatrickDaSilva's answer from almost $4$ years ago, with only some stutter added. Surely there must be better ways to use/waste your time. $\endgroup$ – dxiv Feb 11 '17 at 7:40

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