What is (x+10)^2+(y-5)^2=104 ?

The solution to (x+10)^2+(y-5)^2=104 is Circle with (a,b)=(-10,5),r=sqrt(104)

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(x+10)^2+(y-5)^2=104

(x + 10)^{2} + (y - 5)^{2} = 104

(a, b) = (- 10, 5), r = 104

Solution steps

(x + 10)^{2} + (y - 5)^{2} = 104

Rewrite

(x + 10)^{2} + (y - 5)^{2} = 104

in the form of the standard circle equation

(x - (- 10))^{2} + (y - 5)^{2} = (104)^{2}

Therefore the circle properties are:

(a, b) = (- 10, 5), r = 104

x^{2} + y^{2} < 1

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

x^{2} + y^{2} + 2 x - 3 = 0

x^{2} + y^{2} + 12 y = - 32

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

What is (x+10)^2+(y-5)^2=104 ?
The solution to (x+10)^2+(y-5)^2=104 is Circle with (a,b)=(-10,5),r=sqrt(104)