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

The diameter ((y-2)^2)/9+((x+1/2)^2)/9 =1 is d=6

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

diameter

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

d = 6

Solution steps

d = 2 r

Find the radius

Radius given

\frac{( y - 2 ) ^{2}}{9} + \frac{( x + \frac{1}{2} ) ^{2}}{9} = 1 : r = 3

d = 2 \cdot 3

Refine

d = 6

x^{2} + y^{2} + 8 x - 2 y + 15 = 0

x^{2} + y^{2} = 64

x^{2} - 10 x + y^{2} + 6 y - 1 = 0

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

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

What is the diameter ((y-2)^2)/9+((x+1/2)^2)/9 =1 ?
The diameter ((y-2)^2)/9+((x+1/2)^2)/9 =1 is d=6