Question

Find the coordinates of the circumcentre of the triangle whose vertices are (1,2),(3,-4) and (5,-6). Ans: ( 11,2)

Answer 1

$\boxed{answer}$

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) 2x 2 +y 3−8x−12y+52=x 2 +y 2 −8y+168x+4y=36

2x+y=9 .......(1)

Again, PB 2 =PC 2

(x−0) 2+(y−4) 2=(x−6) 2 +(y−2) 2x 2 +y 2−8y+16=

x 2+y 2 −12x−4y+4012x−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

Explanation:

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