What is the center 3(x+3/4)^2+3(y-1/2)^2=36 ?

The center 3(x+3/4)^2+3(y-1/2)^2=36 is (-3/4 , 1/2)

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

center

3 (x + \frac{3}{4})^{2} + 3 (y - \frac{1}{2})^{2} = 36

(- \frac{3}{4}, \frac{1}{2})

Solution steps

3 (x + \frac{3}{4})^{2} + 3 (y - \frac{1}{2})^{2} = 36

Rewrite

3 (x + \frac{3}{4})^{2} + 3 (y - \frac{1}{2})^{2} = 36

in the form of the standard circle equation

(x - (- \frac{3}{4}))^{2} + (y - \frac{1}{2})^{2} = (23)^{2}

Therefore the circle properties are:

(a, b) = (- \frac{3}{4}, \frac{1}{2}), r = 23

And the center is:

(- \frac{3}{4}, \frac{1}{2})

4 (x - 2)^{2} + 4 y^{2} = 4

(x + 4)^{2} + (y - 1)^{2} = 25

(x - 1)^{2} + (y - 2)^{2} = 1

x^{2} + y^{2} - 2 x - 8 = 0

(x + 11)^{2} + (y + 5)^{2} = 4

What is the center 3(x+3/4)^2+3(y-1/2)^2=36 ?
The center 3(x+3/4)^2+3(y-1/2)^2=36 is (-3/4 , 1/2)