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Beliebt Geometry >

x^2+y^2>= 16

Lösung

x^{2} + y^{2} \geq 16

Lösung

(a, b) = (0, 0), r = 4

Schritte zur Lösung

x^{2} + y^{2} \geq 16

Rewrite

x^{2} + y^{2} \geq 16

in the form of the standard circle equation

(x - 0)^{2} + (y - 0)^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (0, 0), r = 4

Graph

Beliebte Beispiele

x^2+y^2-4x+14y+4=0

x^{2} + y^{2} - 4 x + 14 y + 4 = 0

x^{2} + y^{2} - 1 > 0

2x^2+2y^2-x+y-1=0

2 x^{2} + 2 y^{2} - x + y - 1 = 0

3x^2+3y^2-12x+6y-12=0

3 x^{2} + 3 y^{2} - 12 x + 6 y - 12 = 0

x^2+y^2-4x+6y-17=0

x^{2} + y^{2} - 4 x + 6 y - 17 = 0