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

The solution to 6+x^2+4x+y^2-3y=0 is Circle with (a,b)=(-2, 3/2),r= 1/2

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

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

(a, b) = (- 2, \frac{3}{2}), r = \frac{1}{2}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 2))^{2} + (y - \frac{3}{2})^{2} = (\frac{1}{2})^{2}

Therefore the circle properties are:

(a, b) = (- 2, \frac{3}{2}), r = \frac{1}{2}

x^{2} - 8 x + y^{2} - 10 y = - 5

(x - 5)^{2} + (y + 2)^{2} = 1 0^{2}

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

x^{2} + y^{2} + 4 x + 4 y + 7 = 0

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

What is 6+x^2+4x+y^2-3y=0 ?
The solution to 6+x^2+4x+y^2-3y=0 is Circle with (a,b)=(-2, 3/2),r= 1/2