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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x^{2} + y^{2} - 22 x + 12 y - 150 = 0

(x + 1)^{2} + (y - 5)^{2} = 9

x^{2} + y^{2} + 20 x - 18 y + 145 = 0

2 x^{2} + 8 x + 2 y^{2} + 3 y = 0

x^{2} - 4 x + y^{2} + 6 y = 113

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