What is ((x-5)^2)/1+((y+3)^2)/1 =1 ?

The solution to ((x-5)^2)/1+((y+3)^2)/1 =1 is Circle with (a,b)=(5,-3),r=1

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((x-5)^2)/1+((y+3)^2)/1 =1

\frac{( x - 5 ) ^{2}}{1} + \frac{( y + 3 ) ^{2}}{1} = 1

(a, b) = (5, - 3), r = 1

Solution steps

\frac{( x - 5 ) ^{2}}{1} + \frac{( y + 3 ) ^{2}}{1} = 1

Rewrite

\frac{( x - 5 ) ^{2}}{1} + \frac{( y + 3 ) ^{2}}{1} = 1

in the form of the standard circle equation

(x - 5)^{2} + (y - (- 3))^{2} = 1^{2}

Therefore the circle properties are:

(a, b) = (5, - 3), r = 1

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

x^{2} + y^{2} = 20 - x

x^{2} + y^{2} = 2000

(x - 5)^{2} + (y + 3)^{2} = 7

(x - 9)^{2} + (y + 7)^{2} = 1 1^{2}

What is ((x-5)^2)/1+((y+3)^2)/1 =1 ?
The solution to ((x-5)^2)/1+((y+3)^2)/1 =1 is Circle with (a,b)=(5,-3),r=1