What is a-b+a^2+b^2=0 ?

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

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a-b+a^2+b^2=0

a - b + a^{2} + b^{2} = 0

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

Solution steps

a - b + a^{2} + b^{2} = 0

Rewrite

a - b + a^{2} + b^{2} = 0

in the form of the standard circle equation

(a - (- \frac{1}{2}))^{2} + (b - \frac{1}{2})^{2} = (\frac{1}{2})^{2}

Therefore the circle properties are:

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

2 x^{2} - 4 x + 2 y^{2} + 18 y = 7

(x - 1)^{2} + y^{2} - 1 = \frac{3}{6} x

(x - 6)^{2} + (y - 10)^{2} = 5

2 x^{2} + 2 y^{2} + 8 x - 8 y = 34

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

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