1.

Find the coordinates of the circumcentre of the triangle whose vertices are (3, 0), (-1, - 6) and (4,-1). Also, find its circumradius.

Answer»

Coordinates of points on a circle are A (3, 0), B(-1, - 6) and C(4,-1) 

Let the coordinates of the centre of the circle be O(x, y) 

Using distance formula = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

Since the distance of the points A, B and C will be equal from the center, therefore 

⇒ OA = OC

\(\sqrt{(x - 3)^2 + (y - 0)^2}\) = \(\sqrt{(x + 1)^2 + (y + 6)^2}\)

On squaring both sides, we get

(x - 3)2 + (y - 0)2 = (x + 1)2 + (y + 6)

⇒ x2 + 9 - 6x + y2 = x2 + 2x +1 + y2 + 36 + 12y 

⇒ - 8x - 12y = 28 

⇒ 2x + 3y = -7------------- (1) 

Similarly, OC = OB

\(\sqrt{(x - 4)^2 + (y + 1)^2}\) = \(\sqrt{(x + 1)^2 + (y + 6)^2}\)

On squaring both sides, we get 

(x - 4)2 + (y + 1)2 = (x + 1)2 + (y + 6)

⇒ x2 + 16 - 8x + y2 + 1 + 2y = x2 + 1 +2x + y2 + 36 + 12y 

⇒ - 10x - 10y = 20 

⇒ x + y = - 2 ------------- (2) 

Solving eqn (1) and (2), we get 

x = 1; y = - 3 

Coordinates of circum center are (1, - 3) 

Circum radius of the circle = OA = \(\sqrt{(1 - 3)^2 + (3)^2}\)

\(\sqrt{4 +13}\)

\(\sqrt{13}\) units



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