What is (x^2-2x)+(y^2-2y)=7 ?

The solution to (x^2-2x)+(y^2-2y)=7 is Circle with (a,b)=(1,1),r=3

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(x^2-2x)+(y^2-2y)=7

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

(a, b) = (1, 1), r = 3

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (1, 1), r = 3

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

3 x^{2} - 12 x + 3 y^{2} - 6 y - 12 = 0

x^{2} + y^{2} + 4 x + 10 y + 25 = 0

x^{2} + y^{2} + 2 x - 10 y - 4 = 0

x^{2} + y^{2} \geq 2 x

What is (x^2-2x)+(y^2-2y)=7 ?
The solution to (x^2-2x)+(y^2-2y)=7 is Circle with (a,b)=(1,1),r=3