What is the center (x-3)^2+(y-(-2))^2=3sqrt(5) ?

The center (x-3)^2+(y-(-2))^2=3sqrt(5) is (3,-2)

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center (x-3)^2+(y-(-2))^2=3sqrt(5)

center

(x - 3)^{2} + (y - (- 2))^{2} = 35

(3, - 2)

Solution steps

(x - 3)^{2} + (y - (- 2))^{2} = 35

Rewrite

(x - 3)^{2} + (y - (- 2))^{2} = 35

in the form of the standard circle equation

(x - 3)^{2} + (y - (- 2))^{2} = (35)^{2}

Therefore the circle properties are:

(a, b) = (3, - 2), r = 35

And the center is:

(3, - 2)

(x + 5)^{2} + (y + 6)^{2} = \frac{2}{3}

x^{2} + 5 x + y^{2} - y = \frac{5}{2}

(x - 4)^{2} + (y + 1)^{2} = 18

(x - 4)^{2} + (y + 1)^{2} = 16

x^{2} + y^{2} - 6 x - 12 y - 55 = 0

What is the center (x-3)^2+(y-(-2))^2=3sqrt(5) ?
The center (x-3)^2+(y-(-2))^2=3sqrt(5) is (3,-2)