AP EAMCET Engineering 26 Apr 2017 Shift 1 Paper

Consider the figure shown below.Let O(x,y,z) be the circumcentre of ΔABC.
OA=OB=OC
Thus,
OA=OB
OA2=OB2
(x−1)2+(y−2)2+(z−3)2 =(x−3)2+(y+1)2+(z−5)2
4x−6y+4z−21=0
Similarly,
OB2=OC2
(x−3)2+(y+1)2+(z−5)2 =(x−4)2+(y−0)2+(z+3)2
x+y−8z+5=0
Similarly
OA2=OC2
(x−1)2+(y−2)2+(z−3)2 =(x−4)2+(y−0)2+(z+3)2
6x−4y−12z−11=0
On solving the obtained equations,
x=

7

2

y=−

1

2

z=1
Therefore, the circumcentre of ΔABC is (

7

2

,

−1

2

,1)