Question

Find the circum - centre and circum - radius of the triangle whose vertices are (- 2, 3), (2, - 1) and (4, 0).

Answer 1

Step-by-step explanation:

to find the vertices of circumcenter the formula is (X 1 + X 2 +X 3) by 3, (Y1 + Y2 + Y3 )by 3 let (X1 ,Y1) be (- 2, 3) ,(X 2 y 2) b (2, - 1) and (X3,Y3) be (4,0)

circumcentre :(4/3,2/3 )

circum radius :will be the distance between any of the points and the circumcenter

we can find it out using the distance formula

the distance between( 4, 0) and the circumcenter that we have got (4/ 3, 2 /3 )

radius :√70/3

Answer 2

Let O (x, y) be the circumcentre of the triangle.

Then O A = OB = O C

If we take O A = OB

Using distance formula

O A ² = OB²

⇒ (x + 2) ² + (y - 3) ² = (x - 2) ² + (y + 1) ²

⇒ x² + 4 + 4 x + y² + 9 - 6 y = x²+ 4 - 4 x + y² + 1 + 2 y

After cancelling x² and y² from both sides we get linear equation

8 x - 8 y +8 = 0

Or x – y +1 = 0

⇒ y = x + 1 ……………………(1)

Now consider OC = OB

O C ² = OB²

⇒ (x - 4) ² +(y - 0) ² = (x - 2) ² + (y + 1) ²

⇒ x² + 16 - 8 x + y² = x² + 4 - 4 x + y² + 1 +2 y

After cancelling x² and y² from both sides we get linear equation

- 4 x + 11 - 2 y = 0 ………….(2)

Substituting value of y from (1) equation to equation(2)

- 4 x + 11 - 2 (x + 1) = 0

X = 3/2

Since y = x + 1= 3/2 + 1 = 5/2

So Circumcentre is O (3/2 , 5/2)

Check:

O A^2 = (3/2+2) ² +(5/2-3) ² =49/4 + ¼ = 50/4=25/2

O B^2 = (3/2-2) ² + (5/2+1) ²= ¼ + 49/4 = 50/4 = 25/2

O C^2 = (3/2-4) ² +(5/2-0) ² =25/4 +25/4 = 50/4 =25/2

So clearly O(3/2 , 5/2) is equidistant from A,B and C.

Hence O(3/2,5/2) is cicumcentre.

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