A unique circle with 3 points proof 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
 A: 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,
A: 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!
A: In the algebraic way, you have three points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ such that
$(x_i−a)^2+(y_i−b)^2=c_2,i=1,2,3$
$(x_i−a)^2+(y_i−b)^2=c_2,i=1,2,3$
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,
