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Beliebt Geometry >

x^2+y^2<= 4x+4y

Lösung

x^{2} + y^{2} \leq 4 x + 4 y

Lösung

(a, b) = (2, 2), r = 22

Schritte zur Lösung

x^{2} + y^{2} \leq 4 x + 4 y

Rewrite

x^{2} + y^{2} \leq 4 x + 4 y

in the form of the standard circle equation

(x - 2)^{2} + (y - 2)^{2} = (22)^{2}

Therefore the circle properties are:

(a, b) = (2, 2), r = 22

Graph

Beliebte Beispiele

y^2+x^2+24x+10y+160=0

y^{2} + x^{2} + 24 x + 10 y + 160 = 0

(x+4)^2+(y-2)^2=((8^2))/((sqrt(5)))

(x + 4)^{2} + (y - 2)^{2} = \frac{( 8 ^{2} )}{( 5 )}

(x-1)^2+(y-1)^2=20

(x - 1)^{2} + (y - 1)^{2} = 20

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

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

x^2+y^2-6x-4y-18=0

x^{2} + y^{2} - 6 x - 4 y - 18 = 0