What is the center 3x^2+3y^2-18x-18y+42=0 ?

The center 3x^2+3y^2-18x-18y+42=0 is (3,3)

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center 3x^2+3y^2-18x-18y+42=0

center

3 x^{2} + 3 y^{2} - 18 x - 18 y + 42 = 0

(3, 3)

Solution steps

3 x^{2} + 3 y^{2} - 18 x - 18 y + 42 = 0

Rewrite

3 x^{2} + 3 y^{2} - 18 x - 18 y + 42 = 0

in the form of the standard circle equation

(x - 3)^{2} + (y - 3)^{2} = 2^{2}

Therefore the circle properties are:

(a, b) = (3, 3), r = 2

And the center is:

(3, 3)

(x - 4)^{2} + (y - 1)^{2} = 64

x^{2} + y^{2} - 6 x + 7 y - 16 = 0

(x - 5)^{2} + (y - 1)^{2} = 49

x^{2} + (y - 1)^{2} = 36

x^{2} + 4 x + 10 y + y^{2} - 2 = 0

What is the center 3x^2+3y^2-18x-18y+42=0 ?
The center 3x^2+3y^2-18x-18y+42=0 is (3,3)