How do you find the center and radius for ${(x - 9)}^{2} + {(y + 11)}^{2} = 16$ $(x - 9)^2 + (y + 11)^2 = 16$ ?

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How do you find the center and radius for ${(x - 9)}^{2} + {(y + 11)}^{2} = 16$ $(x - 9)^2 + (y + 11)^2 = 16$ ?

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Answer 1 · 2018-05-28T20:39:40

$centre = (9, - 11), radius = 4$ $"centre "=(9,-11)," radius "=4$

Explanation:

$the standard form of the equation of a circle is$ $"the standard form of the equation of a circle is"$

$color(red)(bar(ul(|color(white)(2/2)color(black)((x-a)^2+(y-b)^2=r^2)color(white)(2/2)|)))$

$"where "(a,b)" are the coordinates of the centre and r"$
$"is the radius"$

$(x-9)^2+(y+11)^2=16" is in standard form"$

$"with "(a,b)=(9,-(-11))" and "r=sqrt16$

$"centre "=(9,11)" and radius "=4$

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How do you find the center and radius for ${(x - 9)}^{2} + {(y + 11)}^{2} = 16$ $(x - 9)^2 + (y + 11)^2 = 16$ ?

1 Answer

Explanation:

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How do you find the center and radius for (x - 9)^2 + (y + 11)^2 = 16(x−9)2+(y+11)2=16(x - 9)^2 + (y + 11)^2 = 16?

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Explanation:

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How do you find the center and radius for ${(x - 9)}^{2} + {(y + 11)}^{2} = 16$ $(x - 9)^2 + (y + 11)^2 = 16$ ?