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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

(x + 4)^{2} + (y - 4)^{2} = 22

x^{2} + y^{2} - 3 x = 15 - 6 y

2 x^{2} + x + 2 y^{2} = 6

z^{2} = 1 - y^{2}

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

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