What is (y+4)^2+(x-3)^2=104 ?

The solution to (y+4)^2+(x-3)^2=104 is Circle with (a,b)=(3,-4),r=sqrt(104)

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(y+4)^2+(x-3)^2=104

(y + 4)^{2} + (x - 3)^{2} = 104

(a, b) = (3, - 4), r = 104

Solution steps

(y + 4)^{2} + (x - 3)^{2} = 104

Rewrite

(y + 4)^{2} + (x - 3)^{2} = 104

in the form of the standard circle equation

(x - 3)^{2} + (y - (- 4))^{2} = (104)^{2}

Therefore the circle properties are:

(a, b) = (3, - 4), r = 104

x^{2} + y^{2} - 4 x + 6 y - 6 = 0

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

(x - 2)^{2} + y^{2} = 100

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

x^{2} + y^{2} + 4 x + 8 y - 36 = 0

What is (y+4)^2+(x-3)^2=104 ?
The solution to (y+4)^2+(x-3)^2=104 is Circle with (a,b)=(3,-4),r=sqrt(104)