What is x^2+y^2-2x+4y+8=7 ?

The solution to x^2+y^2-2x+4y+8=7 is Circle with (a,b)=(1,-2),r=2

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x^2+y^2-2x+4y+8=7

x^{2} + y^{2} - 2 x + 4 y + 8 = 7

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

Solution steps

x^{2} + y^{2} - 2 x + 4 y + 8 = 7

Rewrite

x^{2} + y^{2} - 2 x + 4 y + 8 = 7

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x^{2} + y^{2} + 4 x - 12 y - 3 = 0

2 x^{2} - 8 x + 8 + 2 y^{2} + 4 x + 2 = 25

100 = x^{2} + y^{2} + 4 x + 60

(x + 3)^{2} + (y + 3)^{2} = 25

(x - 8)^{2} + (y - 3)^{2} = 9

What is x^2+y^2-2x+4y+8=7 ?
The solution to x^2+y^2-2x+4y+8=7 is Circle with (a,b)=(1,-2),r=2