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7 November, 12:50

What is the general form of the equation of the given circle with center A? CenterA: (-3,12) Radius:5

x2 + y2 + 6x - 24y - 25 = 0

x2 + y2 - 6x + 24y + 128 = 0

x2 + y2 + 6x - 24y + 128 = 0

x2 + y2 + 6x - 24y + 148 = 0

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  1. 7 November, 14:28
    0
    Standard form of a circle is:

    (x-h) ^2 + (y-k) ^2 = r^2

    where

    (h, k) is the center

    and

    r is the radius

    .

    so we have,

    (x-h) ^2 + (y-k) ^2 = r^2

    put the center (-3,12) and radius 5

    (x - (-3)) ^2 + (y-12) ^2 = 5^2

    (x+3) ^2 + (y-12) ^2 = 25 (general form)

    .

    expand to get "general form"

    x^2 + y^2 + Dx + Ey + F = 0

    (x+3) ^2 + (y-12) ^2 = 25

    (x+3) (x+3) + (y-12) (y-12) = 25

    x^2+6x+9 + y^2-24y+144 = 25

    x^2+y^2+6x-24y+153 = 25

    x^2+y^2+6x-24y+128 = 0 (answer)
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