What is [x-(-2)]^2+(y-3)^2=3^2 ?

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

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[x-(-2)]^2+(y-3)^2=3^2

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

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

y^{2} = 36 - x^{2}

(x - \frac{1}{2})^{2} + (y + \frac{1}{4})^{2} = \frac{64}{25}

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

x^{2} + y^{2} - 21 = 0

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