Find the coordinate of the circumcenter …

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Sia ? 6 years, 5 months ago
Let A(3, 0), B(-1, -6) and C(4, -1) be the given points.
Let O(x, y) be the circumcentre of the triangle.
OA = OB = OC
OA2 = OB2
(x - 3)2 + (y - 0)2 = (x + 1)2 + (y + 6 )2
{tex}\Rightarrow{/tex} x2 + 9 - 6x + y2 = x2 + 1 +2x + y2 + 36 + 12y
{tex}\Rightarrow{/tex} x2 - 6x + y2 - x2 - 2x - y2 - 12y = 1 + 36 - 9
{tex}\Rightarrow{/tex} -8x - 12y = 28
{tex}\Rightarrow{/tex} -2x - 3y = 7
{tex}\Rightarrow{/tex} 2x + 3y = -7 ........(i)
Again,
OB2 = OC2
(x + 1)2 + (y + 6)2 = (x - 4)2 + (y + 1)2
{tex}\Rightarrow{/tex} x2 + 1 + 2x + y2 + 36 +12y = x2 +16 - 8x + y2 + 1 + 2y
{tex}\Rightarrow{/tex} x2 + 2x + y2 + 12y - x2 + 8x + y2 - 2y = 16 + 1 - 1 - 36
10x + 10y = -20
x + y = -2 ....... (ii)
Solving (i) and (ii), we get
x = 1, y = -3
Hence circumcentre of the triangle is (1, -3)
Circumradius {tex}= \sqrt { ( 1 + 1 ) ^ { 2 } + ( - 3 + 6 ) ^ { 2 } }{/tex}
{tex}= \sqrt { ( 2 ) ^ { 2 } + ( 3 ) ^ { 2 } }{/tex}
{tex}= \sqrt { 4 + 9 }{/tex}
{tex}= \sqrt { 13 }{/tex} units.
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