Question

The circumcentre of the triangle with vertices
at (4, 6), (0,4) and (6,2) is​

Answer 1

Answer:

Let the coordinates of the circumcentre of the triangle be P(x, y).Circumcentre of a triangle is equidistant from each of the vertices.

So PA = PB = PC

statement : circumcenter is equidistant from the vertices of triangle ABC

Now

\sqrt{(x-4)^{2}+(y-6)^{2}} = \sqrt{(x-0)^{2} + (y-4)^{2}} = \sqrt{(x-6)^{2}+(y-2)^{2}}

(x−4)

2

+(y−6)

2

=

(x−0)

2

+(y−4)

2

=

(x−6)

2

+(y−2)

2

Now take ,

\begin{lgathered}\sqrt{(x-4)^{2}+(y-6)^{2}} = \sqrt{(x-0)^{2} + (y-4)^{2}}\\\implies x^{2}+16-8x+y^{2}+36-12x = x^{2}+y^{2}+16-8x\\\implies 16+36-12x=16\\\implies 36=12x\\\implies x = 3\end{lgathered}

(x−4)

2

+(y−6)

2

=

(x−0)

2

+(y−4)

2

⟹x

2

+16−8x+y

2

+36−12x=x

2

+y

2

+16−8x

⟹16+36−12x=16

⟹36=12x

⟹x=3

\begin{lgathered}\sqrt{(x-0)^{2} + (y-4)^{2}} = \sqrt{(x-6)^{2}+(y-2)^{2}}\\\implies x^{2}+y^{2}+16-8x = x^{2}+36-12x+y^{2}+4-4y\\\implies 16-8x = 36-12x+4-4y\\\implies 16-8(3)=36-12(3)+4-4y\\\implies -8 = 4-4y\\\implies -8-4 =-4y\\\implies -12 = -4y \\\implies y = 3\end{lgathered}

(x−0)

2

+(y−4)

2

=

(x−6)

2

+(y−2)

2

⟹x

2

+y

2

+16−8x=x

2

+36−12x+y

2

+4−4y

⟹16−8x=36−12x+4−4y

⟹16−8(3)=36−12(3)+4−4y

⟹−8=4−4y

⟹−8−4=−4y

⟹−12=−4y

⟹y=3

Answer 2

let the circumcentre be P(x,y)

then PA =PB

√(x-4)^2+(y-6)^=√(x-0)^2+(y-4)^2

x^2-8x+16+y^2-12y+36=x^2+y^2-8y+16

-8x-4y=-36

2x+y=9

PA=PC

√(x-4)^2+(y-6)^2=√(x-6)^2+(y-2)^2

x^2-8x+16+y^2-12y+36=x^2-12x+36+y^2-4y+4

4x-8y=12

x-2y=-3

on solving 1st and 2nd equation we get : x=3 and y=3

the circumcentre of the given traingle =(3,3)