Answer:
ANSWER
Let A(4,6), B(0,4), C(6,2) be the vertices of the given △ABC.
Let P(x,y) be the circumcentre of △ABC. Then,
PA=PB=PC⟹PA
2
=PB
2
=PC
2
Now, PA
2
=PB
2
(x−4)
2
+(y−6)
2
=(x−0)
2
+(y−4)
2
x
2
+y
2
−8x−12y+52=x
2
+y
2
−8y+16
8x+4y=36
2x+y=9 .......(1)
Again, PB
2
=PC
2
(x−0)
2
+(y−4)
2
=(x−6)
2
+(y−2)
2
x
2
+y
2
−8y+16=x
2
+y
2
−12x−4y+40
12x−4y=24
3x−y=6 .....(2)
Solving equation 1 and 2, we get,
x=3,y=3
Therefore, the coordinates of circumcentre of △ABC are P(3,3).
Circumradius = PA=
(4−3)
2
+(6−3)
2
=
10
units
Hope it is helpful to you