What is (x^2+y^2+6x+3y+1)/4 =0 ?

The solution to (x^2+y^2+6x+3y+1)/4 =0 is Circle with (a,b)=(-3,-3/2),r=(sqrt(41))/2

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(x^2+y^2+6x+3y+1)/4 =0

\frac{x ^{2} + y ^{2} + 6 x + 3 y + 1}{4} = 0

(a, b) = (- 3, - \frac{3}{2}), r = \frac{41}{2}

Solution steps

\frac{x ^{2} + y ^{2} + 6 x + 3 y + 1}{4} = 0

Rewrite

\frac{x ^{2} + y ^{2} + 6 x + 3 y + 1}{4} = 0

in the form of the standard circle equation

(x - (- 3))^{2} + (y - (- \frac{3}{2}))^{2} = (\frac{41}{2})^{2}

Therefore the circle properties are:

(a, b) = (- 3, - \frac{3}{2}), r = \frac{41}{2}

(x + 9)^{2} + (y - 1)^{2} = 25

x^{2} + y^{2} + 16 x - 4 y + 58 = 0

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

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

x^{2} + y^{2} + 8 x - 6 y - 24 = 0

What is (x^2+y^2+6x+3y+1)/4 =0 ?
The solution to (x^2+y^2+6x+3y+1)/4 =0 is Circle with (a,b)=(-3,-3/2),r=(sqrt(41))/2