Question 1208377
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Answer: {{{(x-8/3)^2+(y-11/3)^2=50/9}}}



Explanation


Let
A = (1,2)
B = (3,6)
C = (5,4)
The order of the points does not matter. 
The labels are there to help set up the other items mentioned in this solution page.


The circle you're trying to find is known as a <a href="https://www.mathopenref.com/trianglecircumcircle.html">circumcircle</a> aka circumscribed circle.
The circumcenter is the center of the circumcircle.
The circumcenter is at the intersection of the perpendicular bisectors.
Check out <a href="https://www.algebra.com/algebra/homework/Linear-equations/perpendicular-bisector-example1.lesson">this page</a> to see how to find the equation of the perpendicular bisector of AB.
That page concludes with the equations x+2y=10 and y=(-1/2)x+5.
The equation I'll use is x+2y=10.


Follow a similar process to determine an equation for the perpendicular bisector of BC is x-y=-1.


Solving this system {{{system(x+2y = 10,x-y = -1)}}} will pinpoint the circumcenter. 


Subtract the equations. 
This is to cancel the x terms so you can solve for y.
You should get 3y = 11 which solves to y = 11/3. 
This is the y coordinate of the center of the circle.
Plug y = 11/3 into either x+2y=10 or x-y=-1. Solve for x. You should get x = 8/3.
I'll leave the scratch work for the student to do. 


The center of the circle is located at (h,k) = (8/3,11/3)
h = 8/3 = 2.6667 approximately
k = 11/3 = 3.6667 approximately
For each decimal the '6's go on forever but we have to round at some point.


So we have the center of this circle. 
We now need the radius.


Use the distance formula
{{{d = sqrt( (x[2]-x[1])^2+(y[2]-y[1])^2 )}}} 
to calculate the distance from the center (8/3,11/3) to any of the three points A,B or C.
This will yield the radius r. 
I'll let the student do the scratch work.
You should get {{{r = (5/3)*sqrt(2)}}}
Square both sides to get {{{r^2 = 50/9}}}



Side note: {{{(5/3)*sqrt(2)=2.357023}}} approximately and {{{50/9 = 5.55556}}} approximately
The decimal 5.55556 should have the '5's go on forever, but we have to round at some point.


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Summary:


(h,k) = (8/3, 11/3) is the center
{{{r = (5/3)*sqrt(2)}}} is the radius which has both sides square to {{{r^2 = 50/9}}}


The circle template {{{(x-h)^2+(y-k)^2=r^2}}} updates to the final answer {{{(x-8/3)^2+(y-11/3)^2=50/9}}}
You can verify by using <a href="https://www.geogebra.org/calculator">GeoGebra</a> or <a href="https://www.desmos.com/calculator">Desmos</a> 
There are many other similar tools.


Another way to verify would be to plug the coordinates of points A,B,C into the answer equation. Simplifying everything should lead to the same number on each side.
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