What is the center x^2+y^2+6x-4y+3=0 ?

The center x^2+y^2+6x-4y+3=0 is (-3,2)

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center x^2+y^2+6x-4y+3=0

center

x^{2} + y^{2} + 6 x - 4 y + 3 = 0

(- 3, 2)

Solution steps

x^{2} + y^{2} + 6 x - 4 y + 3 = 0

Rewrite

x^{2} + y^{2} + 6 x - 4 y + 3 = 0

in the form of the standard circle equation

(x - (- 3))^{2} + (y - 2)^{2} = (10)^{2}

Therefore the circle properties are:

(a, b) = (- 3, 2), r = 10

And the center is:

(- 3, 2)

y^{2} = - \frac{1}{4} x

f (x) = x^{2} - 8 x + 7

y^{2} = 16 x

\frac{x ^{2}}{9} - \frac{y ^{2}}{25} = 1

y^{2} = - 24 x