how many circles can pass through three points? (demonstration)

Question

How can I prove that just one circle passes through three points?. I, at some way, asked it in a forum, but no one answered me, so I thought could make a forum. Thanks in advance.

Answer 1

Say you have point $A,B,C$. Consider the fact that the center of the circle $O$ has the same distance from $A,B,C$, then it has to lie on the vertical bisector of both $AB$ and $BC$. Assuming $A,B,C$ not on a same line, there 's only one $O$ possible.

Answer 2

Let $\Delta ABC$ our triangle.

If $O$ is a center of the circle then $OA=OB=OC$, which says that $O$ placed on midperpendiculars to $AB$, to $AC$ and to $BC$, which are unique.

Answer 3

The general equation of a circle is $(x-a)^2+(y-b)^2=r^2$ where $a,b$ refers to it's centre and $r$ is its radius.

Suppose more than one circle pass through three given points. Then, the equation

$(x-a_1)^2+(y-b_1)^2=(x-a_2)^2+(y-b_2)^2$ , must have $3$ solutions which is impossible.

hence, only one circle can pass through three points

Answer 4

As mtheorylord commented, let us write the equations $$(x_1-a)^2+(y_1-b)^2=r^2 \tag 1$$ $$(x_2-a)^2+(y_2-b)^2=r^2 \tag 2$$ $$(x_3-a)^2+(y_3-b)^2=r^2 \tag 3$$ Subtract $(1)$ from $(2)$ and $(1)$ from $(3)$; this gives two linear equations in $a,b$. $$2(x_1-x_2) a+2(y_1-y_2)b=(x_1^2+y_1^2)-(x_2^2+y_2^2) \tag 4$$ $$2(x_1-x_3) a+2(y_1-y_3)b=(x_1^2+y_1^2)-(x_3^2+y_3^2) \tag 5$$ If the solution does exist (I let you finding the conditions - it is easy), it is unique.

Solve $(4)$ and $(5)$ for $a$ and $b$ and replace in $(1)$ to get $r^2$.