Find the center and radius of 2x^2 + 2y^2 -6x - 2y - 13 = 0?

To find the center and radius of the given equation of a circle, we first need to rewrite the equation in standard form, which is:

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

where (h, k) is the center of the circle and r is the radius.

Given equation: 2x^2 + 2y^2 - 6x - 2y - 13 = 0

Rearranging terms, we get:

2x^2 - 6x + 2y^2 - 2y = 13

Dividing by 2 to simplify:

x^2 - 3x + y^2 - y = 6.5

Completing the square for x terms:

(x^2 - 3x + 2.25) + y^2 - y = 6.5 + 2.25

(x - 1.5)^2 + (y - 0.5)^2 = 8.75

Comparing with the standard form, we have:

Center: (1.5, 0.5)
Radius: sqrt(8.75) = 2.958

Therefore, the center of the circle is (1.5, 0.5) and the radius is approximately 2.958.

2x^2 -6x 2y^2 -2y - 13 = 0
(x^2 -3x) +(y^2 - y) = 13/2
(x^2 -3x +9/4) -9/4 +(y^2 - y+1/4) -1/4 = 13/2
(x -3/2)^2 -9/4 +(y -1/2)^2 -1/4 = 13/2
(x -3/2)^2 +(y -1/2)^2 = 13/2 +9/4 +1/4
(x -3/2)^2 +(y -1/2)^2 = 13/2 +10/4
(x -3/2)^2 +(y -1/2)^2 = 23/4

Center =(h,k) = (3/2 ,1/2)
Radius = √(23/4)