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

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

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(x+4)^2+(y-4)^2=2sqrt(2)

(x + 4)^{2} + (y - 4)^{2} = 22

(a, b) = (- 4, 4), r = 2^{\frac{3}{4}}

Solution steps

(x + 4)^{2} + (y - 4)^{2} = 22

Rewrite

(x + 4)^{2} + (y - 4)^{2} = 22

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 4, 4), r = 2^{\frac{3}{4}}

x^{2} + y^{2} - 3 x = 15 - 6 y

2 x^{2} + x + 2 y^{2} = 6

z^{2} = 1 - y^{2}

(x - 9)^{2} + y^{2} = 2

(x + 2)^{2} + (y - 1.5)^{2} = (2.75 i)^{2}

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