Find the co-ordinates of circumcentre and radius of circumcircle of `triangle` ABC if A(7,1) ,B(3,5) and C(2,0) are given .

Explanation

`A(7,1),B(3,5)` and `C(2,0)`
Let `O(a,b)` be the circumcentre.
`OA-OB` ………..(Radii of the same circle)
`:.sqrt((a-7)^(2)+(b-1)^(2))=sqrt((a-3)^(2)+(b-5)^(2))`
………….(Distance formula)
Squaring both the sides we get
`(a-7)^(2)+(b-1)^(2)=(a-3)^(2)+(b-5)^(2)`
`:.a^(2)-14a+49+b^(2)-2b+1`
`=a^(2)-6a+9+b^(2)-10b+25`
`:.-14a-2b+50=-6a-10b+34`
`:.-14a+6a-2b+10b=34+50`
`:.-8a+8b=-16`
`-8(a-b)=-16`
`:.a-b=(-16)/(-8)`
`:.a-b=2`.....1
`OB=OC` .........(Radii of the same circle)
`:.sqrt((a-3)^(2)+(b-5)^(2))=sqrt((a-2)^(2)+(b-0)^(2))`
Squaring both the sides we get,
`(a-3)^(2)+(b-5)^(2)=(a-2)^(2)+(b-10)^(2)`
`:.a^(2)-6a+9+b^(2)-10b+25=a^(2)-4a+4+b^(2)`
`:.-6a-10b+34=-4a+4`
`:.-6a+4a-10b=-4-34`
`:.-2a-10b=-30`
`:.-2a-10b=-30`
`:.a=6b=(-30)/(-2)`
`:.a+5b=15`............2
Subtracting 1 from 2 we get

`:.b=13/6`
Substituting `b=13/6` in 1 we get
`a-b=2`
`:.a-13/6=2`
`:.a=2+13/6`
`:.a=(12+13)/6`
`:.a=25/6`
`:.` coordinates of circumcentre are `(25/6, 13/6)`
By distance formula,
Radius `=OC=sqrt((2-25/6)^(2)+(0-13/6)^(2))`
`=sqrt(((12-25)/6)^(2)+((-13)/6)^(2))`
`=sqrt(((-13)/6)^(2)+((-13)/6)^(2))`
`=sqrt(169/36+169/36)`
`=sqrt((169xx2)/36)`
`=(13sqrt(2))/6`
Radius of the circumcircle is `(13sqrt(2))/6`
Coordinates of circumcentre are `(25/6, 13/6)`
and radius of circumference is `(13sqrt(2))/6`.