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

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

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

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

(a, b) = (2, 4), r = 1

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (2, 4), r = 1

x^{2} + 6 x + y^{2} - 7 = 0

x^{2} + 8 x + y^{2} - 4 y + 16 = 0

x^{2} + (y + 14. 5^{2} - 5^{2})^{2} = 14. 5^{2}

x^{2} + 5 x + y^{2} + 9 y - 10 = 0

x^{2} + y^{2} - 4 x - 8 y + 16 = 0

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