What is ((x-3)^2)/(25)+((y+1)^2)/(25)=1 ?

The solution to ((x-3)^2)/(25)+((y+1)^2)/(25)=1 is Circle with (a,b)=(3,-1),r=5

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((x-3)^2)/(25)+((y+1)^2)/(25)=1

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{25} = 1

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

Solution steps

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{25} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{25} = 1

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x^{2} + y^{2} + 9 y - 2 = 0

x^{2} + y^{2} - 2 x - 4 y - 15 = 0

4 x^{2} + 4 y^{2} - 24 y + 51 = 16 x

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

\frac{x ^{2}}{4} + \frac{y ^{2}}{4} = 25

What is ((x-3)^2)/(25)+((y+1)^2)/(25)=1 ?
The solution to ((x-3)^2)/(25)+((y+1)^2)/(25)=1 is Circle with (a,b)=(3,-1),r=5