What is ((x-5)^2)/4+((y+3)^2)/4 =1 ?

The solution to ((x-5)^2)/4+((y+3)^2)/4 =1 is Circle with (a,b)=(5,-3),r=2

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((x-5)^2)/4+((y+3)^2)/4 =1

\frac{( x - 5 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{4} = 1

(a, b) = (5, - 3), r = 2

Solution steps

\frac{( x - 5 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{4} = 1

Rewrite

\frac{( x - 5 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{4} = 1

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (5, - 3), r = 2

(x - 6)^{2} + (y - 4)^{2} = 8^{2}

(x - 4)^{2} + (k (x) - 3)^{2} = 5^{2}

x^{2} + y^{2} + 6 y - 18 = 0

3 x^{2} + 3 y^{2} + 5 x - 6 y + 1 = 0

x^{2} - 8 x + y^{2} - 10 y + 36 = 0

What is ((x-5)^2)/4+((y+3)^2)/4 =1 ?
The solution to ((x-5)^2)/4+((y+3)^2)/4 =1 is Circle with (a,b)=(5,-3),r=2