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This question already has an answer here:

In Math Olympiad, I needed to create an equation for circle that connects points $A(3;0)$, $B(5;4)$, $C(0;3)$.

I did it in this way:

1) draw segments connecting those points - $AB, BC, CA$

2) draw perpendiculars from the centre of the segments

3) the point where all perpendiculars meet is the centre of circle that connects points $A(3;0)$, $B(5;4)$, $C(0;3)$

4) calculated the Radius of the circle ($R = 2.69$) and created equation: \begin{align*} (x−a)^2 + (y−b)^2 & = r^2\\ (x−2.67)^2 + (y−2.67)^2 & = 7.22 \end{align*}

[img]https://i.imgur.com/J3I40BB.png[/img]

But I am wondering if either is possible:

1) to solve it in easier/faster way

2) to create formula to get a centre of circle that connects points $A(3;0)$, $B(4;5)$, $C(0;3)$.

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marked as duplicate by amd, user99914, Shailesh, Robert Z, Martin R Dec 3 '17 at 9:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ See this question for several methods. Additionally, you know from points $A$ and $C$ that the center must lie somewhere on the line $x=y$. $\endgroup$ – amd Dec 2 '17 at 23:44
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you can plug in the coordinates of Points in the General circle equation: $$(x-x_M)^2+(y-y_M)^2=R^2$$ this gives $$(3-x_M)^2+y_M^2=R^2$$ $$(4-x_M)^2+(5-y_M)^2=R^2$$ $$x_M^2+(3-y_M)^2=R^2$$ and solve this for $$x_M,y_M,R$$

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