What is (x-3)^2+(y-4)^2=1^2 ?

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

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

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

(a, b) = (3, 4), r = 1

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (3, 4), r = 1

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

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

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

x^{2} - 16 x + 64 + y^{2} - 14 x + 49 = 0

5 x^{2} - 5 x + 5 y^{2} - 4 y - 9 = 0

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