Question

the coordinate of the circumcentre of the triangle having vertics (-2,-3) , (-1,0) and (7,-6) are-

Answer 1

Answer:

Centre of circumcircle = (-1/3, 2/3)

Step-by-step explanation:

Accurate Question -

Find the circumcentre of the triangle whose vertices are (-2,-3), (-1,0), (7,-6)

Solution -

Consider the -

O = (a, b) as the co-ordinate of circumcircle of given triangle.

A = (-3,1) , B = (0,-2) and C = (1,3)

Therefore,

AO = BO = CO = radius of circumcircle.

AO² = BO² = CO²

As we know,

Distance Formula -

$\bigstar\;\boxed{\sf(a + 3)^2 + (b - 1)^2 = a^2 + (b + 2)^2 = (a - 1)^2 + (b - 3)^2}}$

Now,

⇒ (a + 3)² + (b - 1)² = a² + (b + 2)²

⇒ a² + 6a + 9 + b² - 2b + 1 = a² + b² + 4b + 4

⇒ 6a - 6b + 6 = 0

⇒ a - b + 1 = 0       ....Eq (1)

Same here,

⇒ a² + (b + 2)² = (a - 1)² + (b - 3)²

⇒ a² + b² + 4b + 4 = a² - 2a + 1 + b² - 6b + 9

⇒ 2a + 10b - 6 = 0

⇒ a + 5b - 3 = 0           ....Eq (2)

From Eq 1 & 2 -

⇒ 5a + a +5 - 3 = 0

⇒ 6a + 2 = 0 , a = $\sf \dfrac{-1}{3}$

⇒ b = $\sf \dfrac{-1}{3}$ + 1 = $\sf \dfrac{2}{3}$

∴ Centre of circumcircle = (-1/3, 2/3)