Q.

Find the circumcenter of the triangle whose  sides are given x+y+2=0,5xy2=0 and x2y+5=0 

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Detailed Solution

Given sides of triangle are
x+y+2=0 ....(1)5xy2=0 ....(2)x2y+5=0 ...(3)
P. I of (1) & (2) is A
x2+2=y10+2=115=x0=y12=16x=0,y=2;A(0,2)
P.I of (1) & (3) is B
x5+4=y25=121=x9=y3=13x=3,y=1;B=(3,1)
P.I of (2) & (3) is C
x54=y225=110+1=x9=y27=19x=1,y=3;C=(1,3)A(0,2),B(3,1),C(1,3)
 Let S(α,β) be the circumcentre of ∆ABC
SA=SB=SCSA=SBSA2=SB2(α0)2+(β+2)2=(α+3)2+(β1)2α2+β2+4+4β=α2+9+6α+β2+12β4+4β96α1+2β=06α+6β6=0αβ+1=0  ...(4)
SB = SC
SB2=SC2(α+3)2+(β1)2=(α1)2+(β3)2α2+9+6α+β2+12β= α2+12α+β2+96β6α+12β1+2α+6β=08α+4β=02α+β=0...(5) 
Solving (4) &(5) we get S(α,β)
α01=β20=11+2α=13,β=23
 ∴Circumcentre of ABC is 13,23

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