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

The solution to x-3y+x^2+y^2=0 is Circle with (a,b)=(-1/2 , 3/2),r=sqrt(5/2)

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

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

(a, b) = (- \frac{1}{2}, \frac{3}{2}), r = \frac{5}{2}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- \frac{1}{2}))^{2} + (y - \frac{3}{2})^{2} = (\frac{5}{2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{1}{2}, \frac{3}{2}), r = \frac{5}{2}

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

x^{2} + y^{2} + 2 y = 15

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

(x + 3)^{2} + (y - 7)^{2} = 58

\frac{( 9 )}{( 2 )} x^{(2)} + \frac{( 9 )}{( 2 )} y^{(2)} = 1

What is x-3y+x^2+y^2=0 ?
The solution to x-3y+x^2+y^2=0 is Circle with (a,b)=(-1/2 , 3/2),r=sqrt(5/2)