Problem: circumcentre of a triangle whose vertices are A(4,1),B(-2,-3),C(6,7) [ Calculator, Method and examples ]
Solution:
Your problem `->` circumcentre of a triangle whose vertices are A(4,1),B(-2,-3),C(6,7)
The center of a circle which passes through all the vertices of a triangle is called circumcentre of a triangle.
Vertices of triangle are `A(4,1),B(-2,-3)` and `C(6,7)`
Let `P(x,y)` be the circumcentre of a triangle
`:. PA=PB=PC`
`:. PA^2=PB^2=PC^2`
Now `PA^2=PB^2`
`(x-4)^2+(y-1)^2=(x+2)^2+(y+3)^2`
`(x^2-8x+16)+(y^2-2y+1)=(x^2+4x+4)+(y^2+6y+9)`
`-12x-8y=-4`
`3x+2y=1 ->(1)`
Now `PB^2=PC^2`
`(x+2)^2+(y+3)^2=(x-6)^2+(y-7)^2`
`(x^2+4x+4)+(y^2+6y+9)=(x^2-12x+36)+(y^2-14y+49)`
`16x+20y=72`
`4x+5y=18 ->(2)`
Intersection point of equation `(1)` and `(2)`
The point of intersection of the lines can be obtainted by solving the given equations
`3x+2y=1`
and `4x+5y=18`
`3x+2y=1 ->(1)`
`4x+5y=18 ->(2)`
equation`(1) xx 5 =>15x+10y=5`
equation`(2) xx 2 =>8x+10y=36`
Substracting `=>7x=-31`
`=>x=-31/7`
`=>x=-4.4286`
Putting `x=-4.4286` in equation `(1)`, we have
`3(-4.4286)+2y=1`
`=>2y=1+13.2857`
`=>2y=14.2857`
`=>y=14.2857/2`
`=>y=7.1429`
`:.x=-4.4286" and "y=7.1429`
`:. (-4.4286,7.1429)` is the intersection point of the given two lines.
`:.` Circumcentre of a triangle is `P(-4.4286,7.1429)`
Radius of the Circumcircle
`AP=sqrt((-4.4286-4)^2+(7.1429-1)^2)=sqrt((-8.4286)^2+(6.1429)^2)=sqrt(71.0413+37.7352)=sqrt(108.7765)=10.4296`
