Question 771838
find the center and radius of the circle passes through (2,3), (6,1) and (4,-3).
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{{{(x-h)^2+(y-k)^2=r^2}}}
solve for h, k, and r^2 with following system of 3 equations:
..
(2,3)   (2-h)^2+(3-k)^2=r^2
(6,1)   (6-h)^2+(1-k)^2)=r^2
(4,-3)  (4-h)^2+(-3-k)^2)=r^2
..
(2-h)^2+(3-k)^2=r^2
(6-h)^2+(1-k)^2=r^2
expand:
4-4h+h^2+9-6k+k^2=r^2
36-12h+h^2+1-2k+k^2=r^2
subtract:
-32+8h+8-4k=0
-24+8h-4k=0
-6+2h-k=0
..
(6-h)^2+(1-k^2)=r^2
(4-h)^2+(-3-k)^2=r^2
expand:
36-12h+h^2+1-2k+k^2=r^2
16-8h+h^2+9+6k+k^2=r^2
subtract:
20-4h-8-8k=0
12-4h-8k=0
3-h-2k=0
..
-6+2h-k=0
3-h-2k=0
..
-12+4h-2k=0
3-h-2k=0
subtract
-15+5h=0
5h=15
h=3
k=-6+2h=0
(2-h)^2+(3-k)^2=r^2
(2-3)^2+(3)^2=r^2
r^2=1+9=10
Equation of given circle:
{{{(x-3)^2+y^2=10}}}
center: (3,0)
radius: √10