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Beliebt Geometry >

x^2+z^2<= 16

Lösung

x^{2} + z^{2} \leq 16

Lösung

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

Schritte zur Lösung

x^{2} + z^{2} \leq 16

Rewrite

x^{2} + z^{2} \leq 16

in the form of the standard circle equation

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

Therefore the circle properties are:

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

Graph

Beliebte Beispiele

x^{2} + y^{2} \leq 3

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

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

x^{(2)}-5x+y^{(2)}+4y=46

x^{(2)} - 5 x + y^{(2)} + 4 y = 46

x^2+y^2+4/3 x-4/3 y+4/9 =0

x^{2} + y^{2} + \frac{4}{3} x - \frac{4}{3} y + \frac{4}{9} = 0

3x^2-9x+3y^2+3y= 39/2

3 x^{2} - 9 x + 3 y^{2} + 3 y = \frac{39}{2}