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

The solution to (x-4)^2+y^2>= x^2+y^2 is Parabola with (h,k)=(0,0),p= 1/4

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

(x - 4)^{2} + y^{2} \geq x^{2} + y^{2}

(h, k) = (0, 0), p = \frac{1}{4}

Solution steps

(x - 4)^{2} + y^{2} \geq x^{2} + y^{2}

Rewrite

(x - 4)^{2} + y^{2} \geq x^{2} + y^{2}

in the standard form

4 \cdot \frac{1}{4} (x - 0) = (y - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 0), p = \frac{1}{4}

2 x^{2} - 12 x + 28 y = 38

8 x^{2} + 2 y + 16 x = 0

3 x^{2} + 5 x + k (x) = 0

(x + 1)^{2} = 12 (y - 0)

y = - x^{2} + 18 x

What is (x-4)^2+y^2>= x^2+y^2 ?
The solution to (x-4)^2+y^2>= x^2+y^2 is Parabola with (h,k)=(0,0),p= 1/4