What is (x^2-2x+1)+(y^2-4y+4)=9 ?

The solution to (x^2-2x+1)+(y^2-4y+4)=9 is Circle with (a,b)=(1,2),r=3

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(x^2-2x+1)+(y^2-4y+4)=9

(x^{2} - 2 x + 1) + (y^{2} - 4 y + 4) = 9

(a, b) = (1, 2), r = 3

Solution steps

x^{2} - 2 x + 1 + (y^{2} - 4 y + 4) = 9

Rewrite

x^{2} - 2 x + 1 + (y^{2} - 4 y + 4) = 9

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (1, 2), r = 3

x^{2} + y^{2} - 4 x + 2 y - 1 = 0

x^{2} + y^{2} + 14 x - 12 y = - 81

(x - 3)^{2} + (y - 4)^{2} = 1^{2}

4 x^{(2)} + 4 y^{(2)} - 24 x + 16 y + 51 = 0

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

What is (x^2-2x+1)+(y^2-4y+4)=9 ?
The solution to (x^2-2x+1)+(y^2-4y+4)=9 is Circle with (a,b)=(1,2),r=3