Find the co-ordinates of the circumcentre of a triangle whose vertices are (-3,1) ,(0,-2) and (1,3) .

Explanation

Let `A(-3,1),B(0,-2),C(1,3)` and the circumference `O(a,b)`.
`OA=OC`………..(Radii of the same circle)
`:.sqrt([(a-(-3)]^(2)+(b-1)^(2))=sqrt((a-1)^(2)+(b-3)^(2))`
……(Distance formula)
Squaring both the sides
`(a+3)^(2)+(b-1)^(2)=(a-1)^(2)+(b-3)^(2)`
`:.a^(2)+6a+9+b^(2)-2b+1`
`=a^(2)-2a+1+b^(2)-6b+9`
`:.6a-2b=-2a-6b`
`:.6a+2a=-6b+2b`
`:.8a=-4b`
`:.7a+4b=0`
`4(2a+b)=0`
`:.2a+b=0/4`
`:2a+b=0` ............1
`OB=OC` ....(Radii of the same circle)
`sqrt((a-0)^(2)+[(b-(-2)]^(2))`
`=sqrt((a-1)^(2)+(b-3)^(2))` ........(Distance formula)
Squaring both the sides
`(a-0)^(2)+(b+2)^(2)=(a-1)^(2)+(b-3)^(2)`
`:.a^(2)+b^(2)+4b+4=a^(2)-2a+1+b^(2)-6b+9`
`:.4b+4=-2a-6b+10`
`:.2a+4b+6b=10-4`
`2a+10b=6`
`:.2(a+5b)=6`
`:.a+6b=6/2`
`:.a+5b=3`
Multiplying 2 with equation 2 we get
`2a+10b=6`............3
Subtracting 1 from 3 we get

`:.b=6/9`
`:.b=2/3`
Substituting `b=2/3` in 1 we get
`2a+b=0`
`:.2a+2/3=0`
`:.a=(-2)/3`
`:.a=(-2)/3xx1/2`
`:.a=(-1)/3`
Coordinates of the circumcentre are `((-1)/3,2/3)`