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Beliebt Geometry >

center x^2+y^2-14x+2y+41=0

Lösung

center

x^{2} + y^{2} - 14 x + 2 y + 41 = 0

Lösung

(7, - 1)

Schritte zur Lösung

x^{2} + y^{2} - 14 x + 2 y + 41 = 0

Rewrite

x^{2} + y^{2} - 14 x + 2 y + 41 = 0

in the form of the standard circle equation

(x - 7)^{2} + (y - (- 1))^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (7, - 1), r = 3

And the center is:

(7, - 1)

Graph

Beliebte Beispiele

scheitelpunkte y=x^2+3

v er t i ces y = x^{2} + 3

foki ((x+2)^2)/9+((y-3)^2)/(16)=1

f oc i \frac{( x + 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1

(x^2)/(64)+(y^2)/(16)=1

\frac{x ^{2}}{64} + \frac{y ^{2}}{16} = 1

asymptoten (x^2)/(36)-(y^2)/(64)=1

a sy m pt o t es \frac{x ^{2}}{36} - \frac{y ^{2}}{64} = 1

x^{2} + y^{2} - 25 = 0