Answer:
Circumcenter = (4,0)
Circumcircle = √5
Step-by-step explanation:
The circumcentre is the point of intersection of the perpendicular bisectors of a triangle. The vertices of the triangle are equidistant to the circumcentre.
Let us assume the coordinate of the circumcentre is at O(x, y). Therefore the distance between the cirmcumcenter and the vertices are:
[tex]AO=\sqrt{(x-2)^2+(y-1)^2} =\sqrt{x^2-4x+4+(y^2-2y+1)}\\=\sqrt{x^2+y^2-4x-2y+5} \\\\BO=\sqrt{(x-5)^2+(y-2)^2} =\sqrt{x^2-10x+25+(y^2-4y+4)}\\=\sqrt{x^2+y^2-10x-4y+29} \\\\CO=\sqrt{(x-3)^2+(y-4)^2} \\=\sqrt{x^2-9x+9+(y^2-8y+16)}=\sqrt{x^2+y^2-9x-8y+25} \\[/tex]
AO = BO, therefore
√(x² + y²-4x-2y+5) = √(x² + y² - 10x - 4y + 29)
x² + y²-4x-2y+5 = x² + y² - 10x - 4y + 29
6x + 2y = 24 (1)
BO = CO
√(x² + y² - 10x - 4y + 29) = √(x² + y² - 9x - 8y + 25)
x² + y² - 10x - 4y + 29 = x² + y² - 9x - 8y + 25
-x + 4y = -4 (2)
Multiply equation 2 by 6 and add to equation 1:
26y = 0
y=0
Put y = 0 in -x + 4y = -4
-x + 4(0) = -4
x = 4
The cicumcenter is at (4, 0)
The radius of the circumcircle = AO = BO = CO. Therefore:
[tex]Radius=AO=\sqrt{(4-2)^2+(0-1)^2} =\sqrt{4+1}=\sqrt{5}[/tex]
