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Beliebt Geometry >

y^2+x^2<= 1

Lösung

y^{2} + x^{2} \leq 1

Lösung

(a, b) = (0, 0), r = 1

Schritte zur Lösung

y^{2} + x^{2} \leq 1

Rewrite

y^{2} + x^{2} \leq 1

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (0, 0), r = 1

Graph

Beliebte Beispiele

x^2+y^2-6x-8y-14=0

x^{2} + y^{2} - 6 x - 8 y - 14 = 0

x^2+y^2+10x+12y-3=0

x^{2} + y^{2} + 10 x + 12 y - 3 = 0

x^2+y^2-6x-4y-7=0

x^{2} + y^{2} - 6 x - 4 y - 7 = 0

y = 9 - x^{2} - y^{2}

x^2+y^2+8x-6y+5=0

x^{2} + y^{2} + 8 x - 6 y + 5 = 0