What is ((y-2)^2+(x+1/2)^2)/9 =1 ?

The solution to ((y-2)^2+(x+1/2)^2)/9 =1 is Circle with (a,b)=(-1/2 ,2),r=3

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((y-2)^2+(x+1/2)^2)/9 =1

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

(a, b) = (- \frac{1}{2}, 2), r = 3

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- \frac{1}{2}, 2), r = 3

1 = 16 - x^{2} - y^{2}

x^{2} + y^{2} - 8 x + 18 y - 3 = 0

x^{2} + y^{2} + 10 x + 8 y + 1 = 0

x^{2} + y^{2} = 11 x + 11 y

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

What is ((y-2)^2+(x+1/2)^2)/9 =1 ?
The solution to ((y-2)^2+(x+1/2)^2)/9 =1 is Circle with (a,b)=(-1/2 ,2),r=3