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Beliebt Geometry >

center 2x^2+2y^2-12x+4y-12=0

Lösung

center

2 x^{2} + 2 y^{2} - 12 x + 4 y - 12 = 0

Lösung

(3, - 1)

Schritte zur Lösung

2 x^{2} + 2 y^{2} - 12 x + 4 y - 12 = 0

Rewrite

2 x^{2} + 2 y^{2} - 12 x + 4 y - 12 = 0

in the form of the standard circle equation

(x - 3)^{2} + (y - (- 1))^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (3, - 1), r = 4

And the center is:

(3, - 1)

Graph

Beliebte Beispiele

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

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

center x^2+y^2+2x-8y-19=0

ce n t er x^{2} + y^{2} + 2 x - 8 y - 19 = 0

center x^2+y^2+4x-2y+1=0

ce n t er x^{2} + y^{2} + 4 x - 2 y + 1 = 0

center 2x^2+32x+2y^2-12=-128

ce n t er 2 x^{2} + 32 x + 2 y^{2} - 12 = - 128

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

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