What is (x+4)^2+(y-3)^2=7 ?

The solution to (x+4)^2+(y-3)^2=7 is Circle with (a,b)=(-4,3),r=sqrt(7)

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(x+4)^2+(y-3)^2=7

(x + 4)^{2} + (y - 3)^{2} = 7

(a, b) = (- 4, 3), r = 7

Solution steps

(x + 4)^{2} + (y - 3)^{2} = 7

Rewrite

(x + 4)^{2} + (y - 3)^{2} = 7

in the form of the standard circle equation

(x - (- 4))^{2} + (y - 3)^{2} = (7)^{2}

Therefore the circle properties are:

(a, b) = (- 4, 3), r = 7

(x - 5)^{2} + (y + 3)^{2} = 16

x^{2} + y^{2} + 4 x + 8 y + 16 = 0

X^{2} + Y^{2} = 16

- 11 = - 4 y^{2} - 4 y - 4 x^{2}

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

What is (x+4)^2+(y-3)^2=7 ?
The solution to (x+4)^2+(y-3)^2=7 is Circle with (a,b)=(-4,3),r=sqrt(7)