What is the center 4x^2-40x+4y^2-32y=5 ?

The center 4x^2-40x+4y^2-32y=5 is (5,4)

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center 4x^2-40x+4y^2-32y=5

center

4 x^{2} - 40 x + 4 y^{2} - 32 y = 5

(5, 4)

Solution steps

4 x^{2} - 40 x + 4 y^{2} - 32 y = 5

Rewrite

4 x^{2} - 40 x + 4 y^{2} - 32 y = 5

in the form of the standard circle equation

(x - 5)^{2} + (y - 4)^{2} = (\frac{13}{2})^{2}

Therefore the circle properties are:

(a, b) = (5, 4), r = \frac{13}{2}

And the center is:

(5, 4)

x^{2} + y^{2} - 2 x + 4 y = 0

x^{2} + y^{2} + 2 x + 4 y - 20 = 0

2 x^{2} + 2 y^{2} + 8 x + 16 y + 14 = 0

x^{2} + y^{2} = - 4 x + 3 y

(x + 4)^{2} + (y - 10)^{2} = 274

What is the center 4x^2-40x+4y^2-32y=5 ?
The center 4x^2-40x+4y^2-32y=5 is (5,4)