Circles — Flashcards

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Use the standard form: (x - h)² + (y - k)² = r². Given: h = 3, k = -4, r = 6. So: (x - 3)² + (y + 4)² = 36. Expand: x² - 6x + 9 + y² + 8y + 16 = 36 → x² + y² - 6x + 8y - 11 = 0. Answer: x² + y² - 6x +…

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Compare with (x - h)² + (y - k)² = r². Here: h = -1, k = 1, r² = 4 → r = 2. So center is (-1, 1), radius is 2.

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Let equation be: x² + y² + 2gx + 2fy + c = 0. Plug in points: (1,0): 1 + 2g + c = 0 → (1) (0,-6): 36 - 12f + c = 0 → (2) (3,4): 9 + 16 + 6g + 8f + c = 0 → 25 + 6g + 8f + c = 0 → (3) From (1): c = -1 -…

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Since it touches x-axis, k = a → center is (h, a), radius = a. Equation: (x - h)² + (y - a)² = a² → x² + y² - 2hx - 2ay + h² = 0. Passes through (1, -2): 1 + 4 - 2h + 4a + h² = 0 → h² - 2h + 4a + 5 = …

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Use diameter form: (x - x₁)(x - x₂) + (y - y₁)(y - y₂) = 0 Here: x₁ = 0, y₁ = 0, x₂ = 2, y₂ = -4 So: (x - 0)(x - 2) + (y - 0)(y + 4) = 0 → x(x - 2) + y(y + 4) = 0 → x² - 2x + y² + 4y = 0 Answer: x² + …

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General form: x = h + r cosθ, y = k + r sinθ Here: h = 1, k = -2, r = 3 So: x = 1 + 3 cosθ, y = -2 + 3 sinθ…

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Use general form x² + y² + 2gx + 2fy + c = 0 when: - The circle passes through given points (to substitute and solve) - You need to find center (-g, -f) and radius √(g² + f² - c) - Comparing two circl…

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The equation x² + y² + 2gx + 2fy + c = 0 represents a circle if: 1. Coefficients of x² and y² are equal (and non-zero) 2. No xy term 3. Radius is real → g² + f² - c > 0 Example: 2x² + 2y² + 4x - 6y + …