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x^2 + y^2 - 6x + 8y = 56, What is the area of a circle whose equation

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x^2 + y^2 - 6x + 8y = 56, What is the area of a circle whose equation [#permalink] New post   19 Mar 2021, 23:06
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\(x^2 + y^2 - 6x + 8y = 56\)

What is the area of a circle whose equation is given above?

A. \(25π\)
B. \(81π\)
C. \(162π\)
D. \(6,561π\)
E. Cannot be determined
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B
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x^2 + y^2 - 6x + 8y = 56, What is the area of a circle whose equation [#permalink] New post   20 Mar 2021, 22:39
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Explanation:

Equation of a circle is \((x-h)^2 + (y-k)^2 = r^2\)
Where, \((h, k)\) are the coordinates of the centre and \(r\) is the radius

\(x^2 + y^2 - 6x + 8y = 56\)

\(x^2 - 6x + y^2 + 8y = 56\)

\(x^2 + 9 - 6x + y^2 + 16+ 8y = 56 + 9 + 16\)

\((x-3)^2 + (y+4)^2 = 81\)

Therefore, Centre \(= (3, -4)\) and \(r = 9\)
Area = \(81π\)

Hence, option B
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Re: x^2 + y^2 - 6x + 8y = 56, What is the area of a circle whose equation [#permalink] New post   19 Sep 2022, 22:31
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Re: x^2 + y^2 - 6x + 8y = 56, What is the area of a circle whose equation [#permalink]
19 Sep 2022, 22:31
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