What is 5=(x+2)^2+(y-3)^2 ?

The solution to 5=(x+2)^2+(y-3)^2 is Circle with (a,b)=(-2,3),r=sqrt(5)

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

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

(a, b) = (- 2, 3), r = 5

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 2, 3), r = 5

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

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

x^{2} + y^{2} = 96

x^{2} + y^{2} = 99

8 x^{2} + 8 y^{2} = 200

What is 5=(x+2)^2+(y-3)^2 ?
The solution to 5=(x+2)^2+(y-3)^2 is Circle with (a,b)=(-2,3),r=sqrt(5)