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Find circumcentre of a triangle whose vertices are A(4,1),B(-2,-3),C(6,7)

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Your problem `->` circumcentre of a triangle whose vertices are A(4,1),B(-2,-3),C(6,7)


`A(4,1), B(-2,-3)` and `C(6,7)` form a triangle.

The center of a circle which passes through the three vertices of a triangle is called circum center.
Let P(x,y) be circum center
`:. 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)`

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`

Putting `x=-31/7 ` in equation `(1)`, we have

`3(-31/7)+2y=1`

`=>2y=1+(93/7)`

`=>2y=(7+93)/7`

`=>2y=100/7`

`=>y=50/7`

`:. x=-31/7" and "y=50/7`

`:. (-4.43,7.14)` is the intersection point of the given two lines.


`:. P(-4.43,7.14)` is circum center.








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