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

The solution to x^2+y^2+6x-4y-4=0 is Circle with (a,b)=(-3,2),r=sqrt(17)

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

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

(a, b) = (- 3, 2), r = 17

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 3))^{2} + (y - 2)^{2} = (17)^{2}

Therefore the circle properties are:

(a, b) = (- 3, 2), r = 17

(x - 4)^{2} + (y + 3)^{2} = \frac{16}{9}

(x - 5)^{2} + (y + 2)^{2} = 61

4 x^{2} + 4 y^{2} + 24 x - 8 y - 9 = 0

x^{2} + 8 x + y^{2} - 4 y + 3 = 0

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

What is x^2+y^2+6x-4y-4=0 ?
The solution to x^2+y^2+6x-4y-4=0 is Circle with (a,b)=(-3,2),r=sqrt(17)