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

The solution to (x+3)^2+y^2=1 is Circle with (a,b)=(-3,0),r=1

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(x+3)^2+y^2=1

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

(a, b) = (- 3, 0), r = 1

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 3))^{2} + (y - 0)^{2} = 1^{2}

Therefore the circle properties are:

(a, b) = (- 3, 0), r = 1

(x - 0)^{2} + (y - 0)^{2} = 7^{2}

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

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

\frac{( 2 x - 2 ) ^{2}}{2} + \frac{( 2 y + 6 ) ^{2}}{2} = \frac{46}{2}

x^{2} + y^{2} - 8 x + 10 y + 32 = 0

What is (x+3)^2+y^2=1 ?
The solution to (x+3)^2+y^2=1 is Circle with (a,b)=(-3,0),r=1