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Given eqn of circle, find the range?

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3 Answers

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  • Pope
    Lv 7
    2 months ago
    Favorite Answer

    Your circle equation is in this general form:

    x² + y² + Dx + Ey + F = 0

    center: (-D/2, -E,2)

    radius = √(D² + E² - 4F) / 2

    For this to be a circle, the radius must be real and positive.

    D² + E² - 4F > 0

    In your case:

    (-6)² + (8)² - 4c > 0

    100 - 4c > 0

    4c < 100

    c < 25

    But that would have to be coupled with the given condition, c ≥ 0. Together, that places c on this interval:

    0 ≤ c < 25

  • 2 months ago

    (x-3)^2 = x^2 - 6x + 9.

    (y+4)^2 = y^2 + 8x + 16.

    So (x-3)^2 + (y+4)^2 

    = x^2 + y^2 - 6x + 8x + 25.

    Therefore, if c = 25, the whole thing collapses to a single point (3,-4).  If c < 25, you can move (c - 25) to the right-hand side, where it will represent r^2 for the circle.  It's OK if C is negative, you'll get a bigger circle.

    If c > 25 the equation cannot be satisfied.

  • rotchm
    Lv 7
    2 months ago

    Complete the square to get

    (x-3)² + (y+4)² - 25 + C = 0.  Convince yourself of this.  Then rewrite as

    (x-3)² + (y+4)²  = 25 - C.  

    For this to be a circle, the RHS must be > 0.

    25 - C > 0. Conclusion?

    Done!

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