3x squared+3y squared-30x+27=0?

Question

Acircle is defined by the following equation:
3x squared+3y sqaured-30x+27=0
What are the coordinates of its center?
Enter as an order pair.

Answer 1

Complete the square for x:

3x² - 30x = 3(x² - 10x) = 3[(x - 5)² - 5²] = 3(x - 5)² - 75

so
3x²+3y²-30x+27=0 becomes

3(x - 5)² - 75 + 3y² + 27 = 0
3(x - 5)² + 3y² = 48 divide by 3
(x - 5)² + y² = 16

=> center (5, 0), radius 4

In general, the center of the circle with equation
(x - A)² + (y - B)² = r²
is (A, B), and the radius is r.

Hope this helps.

..

Answer 2

Okay, so we have 3x^2 + 3y^2 - 30x + 27 = 0

Divide through by 3 to get:

x^2 + y^2 - 10x + 9 = 0

Complete the square in x, and you get:

x^2 - 10x + 25 + y^2 = 16
(x - 5)^2 + y^2 = 4^2
(x - 5)^2 + (y - 0)^2 = 4^2

Remembering that the general equation of a circle is:

(x - h)^2 + (y - k)^2 = r^2,

we can see that this is a circle of radius 4, centered at (5, 0)

Answer 3

3x² + 3y² - 30x + 27 = 0

Get it into standard form:
(3x² - 30x) + (3y²) = -27
(x² - 10x) + (y²) = -9
(x² - 10x + (-10/5)²) + (y²) = (-10/5)² - 9
(x - 10/5)² + y² = 25 - 9
(x - 5)² + y² = 16
(x - 5)² + y² = 4²

The centre is located at the point (5, 0).

Answer 4

You can factor out a 3 to start
x^2 + y^2 -10x +9=0
The circle will be offset on the x-axis. To find out exactly where, we complete the square.
x^2-10x+25 + y^2 =16
and finally
(x-5)^2+y^2 = 4^2
The center is at (5,0)

Answer 5

Do your own homework.