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

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

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

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

(a, b) = (- 1, 2), r = 3

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 1, 2), r = 3

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

25 x^{2} - 9 y^{2} = 225

\frac{x ^{2}}{4} + \frac{y ^{2}}{25} = 1

f (x) = x^{2} + 4 x + 4

x = 4 y^{2}

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