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11 May, 19:57

What are the possible solution sets of the equation x2 + y2 + cx + dy + e = 0?

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  1. 11 May, 22:46
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    Hello:

    x² + y² + cx + dy + e = 0

    (x²+cx) + (y²+dy) + e = 0

    (x² + c/2 x + (c/2) ² - (c/2) ²) + y² + d/2y + (d/2) ² - (d/2) ²) + e = 0

    (x+c/2) ² + (y + d2) ² - (c/2) ² - (d/2) ² + e = 0

    (x+c/2) ² + (y + d/2) ² = (c/2) ² + (d/2) ² - e

    case 1:

    (c/2) ² + (d/2) ² - e < 0 : no solutions because : (x+c/2) ² + (y + d/2) ² ≥ 0

    case 2:

    (c/2) ² + (d/2) ² - e = 0 : (x+c/2) ² + (y + d/2) ² = 0 so : one solution (-c/2, - d/2)

    case 3:

    (c/2) ² + (d/2) ² - e > 0 : infinity solutions
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