Question

find the quadinates of the circumcetre of the traingle whose vertex are (8,6) (8,-2) and (2,-2) also find its circumradius.​

Answer 1

Answer:

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.