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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

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

y^{2} = 169 - x^{2}

x^{2} + y^{2} = \frac{49}{16}

x^{2} + y^{2} - 12 x + 8 y + 36 = 0

x^{2} + y^{2} + 20 x - 12 y + 72 = 0

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