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Beliebt Geometry >

center 2x^2+2y^2+16x-32y+134=0

Lösung

center

2 x^{2} + 2 y^{2} + 16 x - 32 y + 134 = 0

Lösung

(- 4, 8)

Schritte zur Lösung

2 x^{2} + 2 y^{2} + 16 x - 32 y + 134 = 0

Rewrite

2 x^{2} + 2 y^{2} + 16 x - 32 y + 134 = 0

in the form of the standard circle equation

(x - (- 4))^{2} + (y - 8)^{2} = (13)^{2}

Therefore the circle properties are:

(a, b) = (- 4, 8), r = 13

And the center is:

(- 4, 8)

Graph

Beliebte Beispiele

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

ce n t er (x - 3)^{2} + (y + 4)^{2} = 2

center (x-0)^2+(y-0)^2=4

ce n t er (x - 0)^{2} + (y - 0)^{2} = 4

center-20-x^2=-16x-20y+y^2

ce n t er - 20 - x^{2} = - 16 x - 20 y + y^{2}

center x^2+y^2-8x-4y-16=0

ce n t er x^{2} + y^{2} - 8 x - 4 y - 16 = 0

center (x+4)^2+(y-2)^2=121

ce n t er (x + 4)^{2} + (y - 2)^{2} = 121