find the radius and center of the cirles

Question

x²+y² -4x-6y-12=0

 

Answer 1

First you have to complete the square with both the y and the x. Start by grouping together the x's and y's, and you might as well add 12 to both sides to get:

(x² - 4x) + (y² - 6y) = 12

Now complete the square, by dividing the -4 in front of the x by 2, and divide the -6 in front of the y by 2:

(x - 2)² + (y - 3)² = 12 + 4 + 9

When we completed the square, we added a (-2)² and a (-3)² to the left, so we have to add them to the right (the +4 and +9)

(x - 2)² + (y - 3)² = 25

From here you can easily get the center and radius. Use the standard form of a circle:

(x - h)² + (y - k)² = r²

The center is (h, k) and the radius is r.

Answer 2

All you need to do is complete two squares.

x² + y² - 4x - 6y - 12 = 0

x² - 4x + y² - 6y = 12

x² - 4x + 4 + y² - 6y + 9 = 12 + 4 + 9

(x - 2)² + (y - 3)² = 25 = 5²

Thus it has the center at (2,3) and a radius r = 5.

Answer 3

The standard, or center-radius, form equation of a circle is (x - h)² + (y - k)² = r² , (h, k) - coordinate of a center of the circle, and r is a radius.
We will use the formula for perfect square trinomial:  (a ± b)² = a² ± 2ab + b²

x² + y² - 4x - 6y - 12 = 0
Our strategy will be to "force" a perfect square trinomials for "x" and "y"
(x² - 2*2*x + 2²) + (y² - 2*3*y + 3²) - 4 - 9 - 12 = 0
                     ↑                         ↑  
                  + 4                      + 9

(x - 2)² + (y - 3)² - 25 = 0 
(x - 2)² + (y - 3)² = 5² ---> (2, 3) - center of the circle and r = 5