What is (x+2)^2+(y-3)^2=40 ?

The solution to (x+2)^2+(y-3)^2=40 is Circle with (a,b)=(-2,3),r=sqrt(40)

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(x+2)^2+(y-3)^2=40

(x + 2)^{2} + (y - 3)^{2} = 40

(a, b) = (- 2, 3), r = 40

Solution steps

(x + 2)^{2} + (y - 3)^{2} = 40

Rewrite

(x + 2)^{2} + (y - 3)^{2} = 40

in the form of the standard circle equation

(x - (- 2))^{2} + (y - 3)^{2} = (40)^{2}

Therefore the circle properties are:

(a, b) = (- 2, 3), r = 40

(x + 2)^{2} + (y - 3)^{2} = 20

(x + 1)^{2} + (y + 6)^{2} = 8^{2}

(x + 2)^{2} + (y - 3)^{2} = 10

(x - 7)^{(2)} + (y - 7)^{(2)} = \frac{( 4 )}{( 9 )}

(x - 2)^{2} + (y - 4)^{2} = 49

What is (x+2)^2+(y-3)^2=40 ?
The solution to (x+2)^2+(y-3)^2=40 is Circle with (a,b)=(-2,3),r=sqrt(40)