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12 January, 11:44

What is the radius of a circle whose equation is x2+y2+8x-6y+21=0?

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Answers (2)
  1. 12 January, 11:46
    0
    Radius=2

    Center: (-4,3)

    Step-by-step explanation:

    The general equation of a circle is

    (x-xo) ² + (y-yo) ²=r², where (xo, yo) are the coordinates of the center, and r is the radius.

    Now: x²+y²+8x-6y+21=0

    x²+8x+y²-6y+21=0

    (x²+8x+16-16) + (y²-6y+9-9) + 21=0

    (x²+8x+16) - 16 + (y²-6y+9) - 9+21=0

    (x+4) ² + (y-3) ²-16-9+21=0

    (x+4) ² + (y-3) ²=16+9-21=25-21=4=2²

    So

    (x+4) ² + (y-3) ²=2².

    Thus, the radius is 2.

    The center is at (-4,3)
  2. 12 January, 12:04
    0
    the radius is 2
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