1.

Find the centre of the circle passing through (2, 1), (5, - 8) and (2, - 9).

Answer»

Coordinates of points on a circle are A(2,1), B(5,-8) and C(2,-9). 

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 = OB

\(\sqrt{(x - 2)^2 + (y - 1)^2}\) = \(\sqrt{(x - 5)^2 + (y + 8)^2}\)

On squaring both sides, we get 

⇒ x2 + 4 - 4x + y2 + 1 - 2y = x2 + 25 - 10x + y2 + 64 + 16y 

⇒ 6x - 18y - 84 = 0 

⇒ x - 3y - 14 = 0 ------------- (1) 

Similarly, OC = OB

\(\sqrt{(x - 2)^2 + (y + 9)^2}\) = \(\sqrt{(x - 5)^2 + (y + 8)^2}\)

⇒ x2 + 4 - 4x + y2 + 81 + 18y 

= x2 + 25 - 10x + y2 + 64 + 16y 

⇒ 6x - 2y - 4 = 0 

⇒ 3x - y - 2 = 0 ------------- (2) 

By solving equations (1) and (2), we get x = -1, y = -5 

So, 

the coordinates of the centre of the circle is (-1, -5). 

Radius of the circle = OA = \(\sqrt{(-1 - 2)^2 + (-5 - 1)^2}\)

\(\sqrt{9+36}\)

\(\sqrt{45}\)

\(3\sqrt5\) units



Discussion

No Comment Found

Related InterviewSolutions