Question

Help asap!! find the center of:

9x^2+y^2-18x-6y+9=0
show work pls!!

Answer 1

center, = 9, 3

radius = 9

Explanation:

9x² + y² - 18x - 6y + 9 = 0

equation of a circle is,

x² + y² + 2ax + 2by + c = 0

where center of a circle equals, -a, -b

radius = √a² + b² - c

by comparing the general equation from the given equation,

2ax = - 18x

a = -9

2by = -6y

b = -3

center of a circle -a, -b will be 9,3

radius = √81 + 9 -9

=√81

=9

Answer 2

To find the center of the given ellipse, we need to first put the equation in standard form:

9x^2 + y^2 - 18x - 6y + 9 = 0

We can start by completing the square for both the x and y terms. For the x terms, we can add and subtract (18/2)^2 = 81 to get:

9(x^2 - 2x + 81/9) + y^2 - 6y + 9 = 0

Simplifying inside the parentheses, we get:

9(x - 9/3)^2 + y^2 - 6y + 9 = 0

For the y terms, we can add and subtract (6/2)^2 = 9 to get:

9(x - 3)^2 + (y - 3)^2 = 36

Dividing both sides by 36, we get:

[(x - 3)^2]/4 + [(y - 3)^2]/36 = 1

Comparing this to the standard form of an ellipse:

[(x - h)^2]/a^2 + [(y - k)^2]/b^2 = 1

We can see that the center of the ellipse is at the point (h, k), which in this case is (3, 3). Therefore, the center of the given ellipse is (3, 3).

Explanation: