Question

01. The vertices A,B,C of a triangle are (2, 1), (5,2)
and (3, 4) respectively. Then the coordinates of
the circum center is​

Answer 1

A basic concept of circumcircle of a triangle is that distance from vertex to circumcenter always remains constant.

i mean, if P (x, y) is circumcenter then,

AP = BP = CP

from distance formula,

AP = √{(x - 2)² + (y - 1)²}

BP = √{(x - 5)² + (y - 2)²}

CP = √{(x - 3)² + (y - 4)²}

case 1 : AP = BP

or, AP² = BP²

or, (x - 2)² + (y - 1)² = (x - 5)² + (y -2)²

or, -4x + 4 - 2y + 1 = -10x + 25 - 4y + 4

or, 6x + 2y = 24 ......(1)

case 2 : BP = CP

or, BP² = CP²

or, (x - 5)² + (y - 2)² = (x - 3)² + (y - 4)²

or, -10x + 25 - 4y + 4 = -6x + 9 - 8y + 16

or, -4x + 4y = -4

or, x - y = 1 ......(2)

from equations (1) and (2),

8x = 26 => x = 13/4

and y = 9/4

hence, circumcenter, P = (13/4, 9/4)

Answer 2

Answer:

(26/7, 6/7)

Explanation:

Given that A (2,1), B (5,2) and C (3,4), then the coordinates of the circum-center are given by:  

(x-h)2 +(y-k)2= r2 where h and k are constants and r is the radius of the circum-circle.

At point A (2,1)⇒ (2-h)2 +(1-k)2= r2…………………………………. (1)

4-4h+h2+ 1-2k+k2

At point B (5,2) ⇒ (5-h)2 +(2-k)2= r2 …………………………………. (2)

25-10h+ h2+ 4-4k+k2

Equating (1) and (2) we get:

25-10h+ h2+ 4-4k+k2=4-4h+h2+ 1-2k+k2

6h+2k=24

3h+k=12 …………………………………………………………………………. (3)

At point C (3,4) ⇒ (3-h)2 +(4-k)2= r2 …………………………………... (4)

9-6h+h2 + 16-8k+ k2= r2

Equating (1) and (4) we obtain:

4-4h+h2+ 1-2k+k2= 9-6h+h2 + 16-8k+ k2

4h+6k=20

2h+3k=10 ……………………………………………………………………… (5)

Solving equation (3) and (5) simultaneously, we have

3h+k=12………………………(i)

2h+3k=10 ……………………(ii)

By substitution method; making k the subject of the formula in (i)

K= 12-3h

Replacing in (ii) we obtain

2h+ 3(12-3h) = 10

7h=26

h= 26/7

k=6/7

hence the coordinates of the circum-center become (26/7, 6/7)