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Beliebt Geometry >

a^2+b^2<= 2a+1

Lösung

a^{2} + b^{2} \leq 2 a + 1

Lösung

(a, b) = (1, 0), r = 2

Schritte zur Lösung

a^{2} + b^{2} \leq 2 a + 1

Rewrite

a^{2} + b^{2} \leq 2 a + 1

in the form of the standard circle equation

(a - 1)^{2} + (b - 0)^{2} = (2)^{2}

Therefore the circle properties are:

(a, b) = (1, 0), r = 2

Graph

Beliebte Beispiele

(x-7)^2+(y-7)^2= 4/9

(x - 7)^{2} + (y - 7)^{2} = \frac{4}{9}

x^2+y^2-10x+6y+18=0

x^{2} + y^{2} - 10 x + 6 y + 18 = 0

(x+3)^2+(y-4)^2=49

(x + 3)^{2} + (y - 4)^{2} = 49

x^2+y^2-6x+2y+1=0

x^{2} + y^{2} - 6 x + 2 y + 1 = 0

x^2+y^2-4x+10y+23=0

x^{2} + y^{2} - 4 x + 10 y + 23 = 0