Question

Find the circumcentre of the triangle whose vertices are (2,1),(5,2)and (3,4)

Answer 1

Answer:

(13/4, 9/4)

Step-by-step explanation:

Hi,

Let the O = (a, b) is the co-ordinate of center of circumcircle of given

triangle.

Let A = (2, 1) , B = (5, 2) and C = (3, 4)

then,

AO = BO = CO = radius of circumcircle.

so, AO² = BO² = CO²

Using distance formula,

(a - 2)² + (b - 1)² = (a - 5)² + (b - 2)² = (a - 3)² + (b - 4)²

now, equating (a - 2)² + (b - 1)² = (a - 5)² + (b - 2)², we get

a² -4a + 4 + b² - 2b + 1 = a² - 10a + 25 + b² - 4b + 4

6a + 2b - 24 = 0

3a + b - 12 = 0 ------(1)

Similarly, (a - 5)² + (b - 2)² = (a - 3)² + (b - 4)²

a² - 10a + 25 + b² - 4b + 4  = a² - 6a + 9 + b² - 8b + 16

4a - 4b - 4 = 0

a - b - 1 = 0 ------(2)

from equations (1) + (2),

4a - 13 = 0

a = 13/4

and b = 9/4

hence, centre of circumcircle = (13/4, 9/4)

Hope, it helps !

Answer 2

Answer:

I hope it helps !!

Step-by-step explanation:

Let the O = (a, b) is the co-ordinate of center of circumcircle of given

triangle.

Let A = (2, 1) , B = (5, 2) and C = (3, 4)

then,

AO = BO = CO = radius of circumcircle.

so, AO² = BO² = CO²

Using distance formula,

(a - 2)² + (b - 1)² = (a - 5)² + (b - 2)² = (a - 3)² + (b - 4)²

now, equating (a - 2)² + (b - 1)² = (a - 5)² + (b - 2)², we get

a² -4a + 4 + b² - 2b + 1 = a² - 10a + 25 + b² - 4b + 4

6a + 2b - 24 = 0

3a + b - 12 = 0 ------(1)

Similarly, (a - 5)² + (b - 2)² = (a - 3)² + (b - 4)²

a² - 10a + 25 + b² - 4b + 4  = a² - 6a + 9 + b² - 8b + 16

4a - 4b - 4 = 0

a - b - 1 = 0 ------(2)

from equations (1) + (2),

4a - 13 = 0

a = 13/4

and b = 9/4

hence, the centre of circumcircle = (13/4, 9/4)