Question

find the coordinates of the circumcentre of the triangle whose vertices (8,6) , (8,-2) & (2,-2). also find the circumradius.

Answer 1

vertices:
A= (8,6)
B= (8,-2)
C= (2, -2)

circumcentre: P(x,y)
circum radii: PA, PB, PC

PA²= PB² = PC²
(x-8)² + (y - 6)² = (x-8)² + (y+2)²
y² -12y +36 = y²+ 4y + 4
y = 2

in a same way x= 5

circumradius = √(5-8)² + (2-6)²
= √9+16
=5

Answer 2

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To Find :- The circum - radius.

Let, A ( 8,6 ) , B ( 8,-2 ) and C ( 2,-2 ) be the vertices of the given triangle.

And also let P ( x,y ) be the circumcenter of this triangle.

PA = PB = PC

PA² = PB² = PC²

Now, PA² = PB²

( x-8 )² + ( y-6 )² = ( x-8 )² + ( y+2 )²

x² + y² - 16x - 12y + 100 = x² + y² - 16x + 4y + 68

16y = 32

y = 2

And

PB² = PC²

( x - 8 )² + ( y + 2 )² = ( x - 2 )² + ( y + 2 )²

x² + y² - 16x + 4y + 68 = x² + y² - 4x + 4y + 8

12x = 60

x = 5

So, the coordinates of the circumcentre P are ( 5,2 )

Also Circum-radius = PA = PB = PC = √(5-8)² + (2-6)²

⇒ 5.