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The circle given by x^2+y^2-6y-12=0 can be written in standard form like this: x^2+(y-k)^2=21 What is the value of k in this equation?

The circle given by x2y26y120 can be written in standard form like this x2yk221 What is the value of k in this equation class=

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absor201

Answer:

The value of k =3

Step-by-step explanation:

[tex]x^2+y^2-6y-12=0[/tex] We need to solve this equation to become in standard form to find the value of k.

[tex]x^2+y^2-6y-12=0\\x^2+y^2-6y = 12\\(x)^2 +(y^2-2(y)(3)+(3)^2) = 12 +(3)^2\\(x)^2+(y-3)^2 = 12+9\\(x)^2 +(y-3)^2 = 21[/tex]

The given standard form is:

[tex]x^2+(y-k)^2=21[/tex]

Comparing it with [tex](x)^2 +(y-3)^2 = 21[/tex]

The value of k =3

Answer:

k = 3

Step-by-step explanation:

Equation of the given function has been given as x² + y² - 6y - 12 = 0.

We have to convert this equation in the standard form of x² + (y - k)² = 21 to get the value of k.

x² + y² - 6y - 12 = 0

x² + y² - 6y + 9 - 9 - 12 = 0

x² + [y²- 2(3y) + 3²] - 21 = 0

x² + (y - 3)²- 21 = 0

x² + (y - 3)² = 21

Now by comparing this equation with the standard form of the equation we get the value of k = 3

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