Question

the circumradius of the triangle formed by the points (4,6),(0,4)and (6,2) is 1) root 7 2) root 8 3)root 10 4) root 11 give reson

Answer 1

Answer:

The circumradius of the triangle is 10 units.

Step-by-step explanation:

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⟹PA² =PB² =PC²

Now, PA² =PB²

(x−4)² +(y−6)² =(x−0)² +(y−4)²

x² +y² −8x−12y+52=x² +y²−8y+16

8x+4y=36

2x+y=9         .......(1)

Again, PB²=PC²

(x−0)² +(y−4)² =(x−6)² +(y−2)²

x² +y² −8y+16=x² +y² −12x−4y+40

12x−4y=24

3x−y=6        .....(2)

Solving equation 1 and 2, we get,

x=3,y=3

The coordinates of circumcentre of △ABC are P(3,3).

Circumradius = PA= (4−3)² +(6−3)² = 10units