Question

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?

Answer 1

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

Answer 2

Answer:

The value of k =3

Step-by-step explanation:

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

$x^2+y^2-6y-12=0\\x^2+y^2-6y = 12\\(x)^2 +(y^2-2(y)(3)+(3)^2) = 12 +(3)^2\\(x)^2+(y-3)^2 = 12+9\\(x)^2 +(y-3)^2 = 21$

The given standard form is:

$x^2+(y-k)^2=21$

Comparing it with $(x)^2 +(y-3)^2 = 21$

The value of k =3