What is the center 4x^2-8x+4y^2+12y-243=0 ?

The center 4x^2-8x+4y^2+12y-243=0 is (1,-3/2)

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center 4x^2-8x+4y^2+12y-243=0

center

4 x^{2} - 8 x + 4 y^{2} + 12 y - 243 = 0

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

Solution steps

4 x^{2} - 8 x + 4 y^{2} + 12 y - 243 = 0

Rewrite

4 x^{2} - 8 x + 4 y^{2} + 12 y - 243 = 0

in the form of the standard circle equation

(x - 1)^{2} + (y - (- \frac{3}{2}))^{2} = 8^{2}

Therefore the circle properties are:

(a, b) = (1, - \frac{3}{2}), r = 8

And the center is:

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

2 x^{2} + 2 y^{2} + 20 x + 16 y + 9 = 0

x^{2} + (y + 20)^{2} = 100

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

(x + 3)^{2} + (y + 5)^{2} = 4

(x + 3)^{2} + (y + 5)^{2} = 9

What is the center 4x^2-8x+4y^2+12y-243=0 ?
The center 4x^2-8x+4y^2+12y-243=0 is (1,-3/2)