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Question

Find the coordinates of the circumcentre of the triangle whose vertices are (3, 0), (-1, -6) and (4, -1). Also , find its circumradius.

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Solution

A(3,0)
B(-1,-6)
C(4,-1)

Let the circumcentre be O(x,y)
Circumcentre is equidistant from A, B and C.
Hence, OA = OB = OC
OA = square root of left parenthesis x minus 3 right parenthesis squared plus y squared end root space space minus negative negative negative negative left parenthesis i right parenthesis
OB = square root of left parenthesis x plus 1 right parenthesis squared plus left parenthesis y plus 6 right parenthesis squared end root space space space minus negative negative negative negative thin space left parenthesis i i right parenthesis
OC = square root of left parenthesis 4 minus x right parenthesis squared plus left parenthesis negative 1 minus y right parenthesis squared end root ------- (iii)

From i and ii
square root of left parenthesis x minus 3 right parenthesis squared plus y squared end root equals square root of left parenthesis x plus 1 right parenthesis squared plus left parenthesis y plus 6 right parenthesis squared end root x squared plus 9 minus 6 x plus y squared equals x squared plus y squared plus 1 plus 36 plus 2 x plus 12 y 9 minus 6 x space equals space 37 space plus 2 x space plus 12 y 8 x plus 12 y space plus 28 space equals 0 2 x plus 3 y plus 7 equals 0
---------(iv)

From (i) and iii

square root of left parenthesis 4 minus x right parenthesis squared plus left parenthesis negative 1 minus y right parenthesis squared end root = square root of left parenthesis x minus 3 right parenthesis squared plus y squared end root
16 plus x squared minus 8 x plus 1 plus y squared plus 2 y equals x squared plus 9 minus 6 x plus y squared 16 minus 8 x plus 1 plus 2 y space equals space 9 minus 6 x 2 y space minus 2 x space plus 8 equals 0 y minus x plus 4 equals 0 space space space space minus negative negative negative space left parenthesis v right parenthesis

from (iv) and 2 cross times (v)
2x+3y+7=0
-2x+2y +8 =0
___________
5y+15=0

y = -3

x = 1

So the circumcenter is (1,-3).
Circumradius =
square root of left parenthesis 1 minus 3 right parenthesis squared plus left parenthesis negative 3 right parenthesis squared end root square root of 4 plus 9 end root square root of 13

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