What is x^{(2)}+y^{(2)}+2x+4y-15=0 ?

The solution to x^{(2)}+y^{(2)}+2x+4y-15=0 is Circle with (a,b)=(-1,-2),r=2sqrt(5)

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x^{(2)}+y^{(2)}+2x+4y-15=0

x^{(2)} + y^{(2)} + 2 x + 4 y - 15 = 0

(a, b) = (- 1, - 2), r = 25

Solution steps

x^{2} + y^{2} + 2 x + 4 y - 15 = 0

Rewrite

x^{2} + y^{2} + 2 x + 4 y - 15 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 1, - 2), r = 25

(x - 3)^{2} + (y - 5)^{2} = 306.25

x^{2} + y^{2} - 8 x + 2 y + 10 = 0

(x + 1)^{2} + (y - 5)^{2} = 27

x^{2} + y^{2} - 10 x + 12 y + 45 = 0

yy = 2 x - x^{2}

What is x^{(2)}+y^{(2)}+2x+4y-15=0 ?
The solution to x^{(2)}+y^{(2)}+2x+4y-15=0 is Circle with (a,b)=(-1,-2),r=2sqrt(5)