Help with math plz?

Question

Write the given equation in the form (x-h)^2+(y-k)^2=r^2, give the center and radius of the circle by the equation.
x^2+y^2+4x-12y-9=0

Answer 1

it's just completing the square. here:
x^2+y^2+4x-12y-9=0
(x^2 + 4x + 4) + (y^2 -12y + 36) = 9+4+36
(x+2)^2 + (y-6)^2 = 49

so center is at (-2, 6) and radius is 7 :)

Answer 2

x^2 + y^2 + 4x - 12y - 9 = 0

x^2 + 4x + y^2 - 12y = 9

add 4 and 36 on both sides to complete the squares

x^2 + 4x + 4 + y^2 - 12y + 36 = 9 + 4 + 36

(x + 2)^2 + (y - 6)^2 = 49

(x + 2)^2 + (y - 6)^2 = 7^2

so center is at (-2, 6) and radius = 7

Answer 3

It helps if you isolate the variables.

For example

put it as x^2+4x+y^2-12y-9=0

then move the constant over to the other side

x^2+4x+y^2-12y=9

then complete the square for each variable

that is what does the 4 in the 4x need to be a (x+z)^2 equation

x^2+4x+4+ y^2-12y+36=9+4+36 (if you add one side you have to add it to another)

thats (x+2)^2+(y-6)^2=49

where h is -2, k is 6, r is 7

so the center is (-2,6) and the radius is 7

Answer 4

x^2 +4x + 4 +y^2 -12y +36 - 9 = 40
(x+2)^2 + (y-6)^2 = 49
So center at (-2,6) and radius = 7

Answer 5

(x-h)^2 + (y-k)^2 -r^2 = x^2 +y^2 +4x -12y -9

x^2-2xh +h^2 +y^2 -2yk +k^2 -r^2 =x^2 + y^2 +4x -12y -9

-2xh +h^2 -2yk +k^2 -r^2 = +4x -12y -9

-2xh= 4x -2yk = -12y
-2h = 4 -2k = -12
h = -2 k = 6

h^2 + k^2 -r^2 =-9

substituting

(-2)^2 + (6)^2 - r^2 = -9

4+ 36 -r^2 = - 9
r^2 = -9 -40 = -49

(x+4)^2 + (y-6)^2 =- 49

Answer 6

(x + 2)^2 + (y - 6)^2 = r^2
x^2 + 4x + 4 + y^2 - 12y + 36 = R^2

So 36 - r^2 = -9
r^2 = 36 + 9 =45
r^2 = 45
r = sq rt(45)

Centre is (-2,6)
radius is sq rt(45)
r^2 = 45

(x + 2)^2 + (y - 6)^2 = r^2
(x + 2)^2 + (y - 6)^2 = 45

OK!!!!