InterviewSolution
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InterviewSolution
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Find the centre of the circle passing through (6, - 6), (3, - 7) and (3, 3). |
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Answer» Coordinates of points on a circle are A(6, -6), B(3, -7) and C(3, 3). 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 \(\sqrt{(x - 6)^2 + (y + 6)^2}\) = \(\sqrt{(x - 3)^2 + (y - 3)^2}\) On squaring both sides, we get (x - 6)2 + (y + 6)2 = (x - 3)2 + (y - 3)2 x2 - 12x + 36 + y2 + 12y + 36 = x2 - 6x + 9 + y2 - 6y + 9 x - 3y = 9........(1) Similarly, OA = OB \(\sqrt{(x - 6)^2 + (y + 6)^2}\) = \(\sqrt{(x - 3)^2 + (y + 7)^2}\) On squaring both sides, we get (x - 6)2 + (y + 6)2 = (x - 3)2 + (y + 7)2 x2 - 12x + 36 + y2 + 12y + 36 = x2 - 6x + 9 + y2 + 14y + 49 3x + y + 7.......(1) Solving eqn (1) and (2), we get x = 3; y = - 2 Coordinates of circum center are (3, - 2) |
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