InterviewSolution
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Find the equation of the circle passing through the points (0, 1), (4, 3) and (1, -1). |
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Answer» Let the required of the circle be x2 + y2 + 2gx + 2fy + c = 0 … (1) It passes through (0, 1) 0 + 1 + 2g(0) + 2f(1) + c = 0 1 + 2f + c = 0 2f + c = -1 … (2) Again the circle (1) passes through (4, 3) 42 + 32 + 2g(4) + 2f(3) + c = 0 16 + 9 + 8g + 6f + c = 0 8g + 6f + c = -25 … (3) Again the circle (1) passes through (1, -1) 12 + (-1)2 + 2g(1) + 2f(-1) + c = 0 1 + 1 + 2g – 2f + c = 0 2g – 2f + c = -2 … (4) 8g + 6f + c = -25 (4) x 4 subtracting we get, 8g – 8f + 4c = -8 14f – 3c = -17 … (5) 14f – 3c = -17 (2) x 3 ⇒ 6f + 3c = -3 Adding we get 20f = -20 f = -1 Using f = -1 in (2) we get, 2(-1) + c = -1 c = -1 + 2 c = 1 Using f = -1, c = 1 in (3) we get 8g + 6(-1) + 1 = -25 8g – 6 + 1 = -25 8g – 5 = -25 8g = -20 g = \(\frac{-20}{8}\) = \(\frac{-5}{2}\) Using g = \(\frac{-5}{2}\), f = -1, c = 1 in (1) we get the equation of the circle. x2 + y2 + 2(\(\frac{-5}{2}\))x + 2(-1)y + 1 = 0 x2 + y2 – 5x – 2y + 1 = 0 |
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