What is (sqrt(x^2+y^2-1))^2>9 ?

The solution to (sqrt(x^2+y^2-1))^2>9 is Circle with (a,b)=(0,0),r=sqrt(10)

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(sqrt(x^2+y^2-1))^2>9

(x^{2} + y^{2} - 1)^{2} > 9

(a, b) = (0, 0), r = 10

Solution steps

(x^{2} + y^{2} - 1)^{2} > 9

Rewrite

x^{2} + y^{2} - 1^{2} > 9

in the form of the standard circle equation

(x - 0)^{2} + (y - 0)^{2} = (10)^{2}

Therefore the circle properties are:

(a, b) = (0, 0), r = 10

9 x^{2} - 6 x + 9 + 9 y^{2} + 24 y + 144 = 19 + 9 + 144

(- 1 - x)^{2} + (2 - y)^{2} = 25

(x - 15)^{2} + (y + 7)^{2} = 7^{2}

15 x^{2} + 15 y^{2} = 60

x^{2} + y^{2} - 6 x + \frac{80}{9} = 0

What is (sqrt(x^2+y^2-1))^2>9 ?
The solution to (sqrt(x^2+y^2-1))^2>9 is Circle with (a,b)=(0,0),r=sqrt(10)