emmyfernandez6 emmyfernandez6
  • 18-07-2019
  • Mathematics
contestada

what is center of the circle given by its equation (x-2)^2 + (y+4)^2=6?

Respuesta :

freckledspots freckledspots
  • 18-07-2019

Answer:

(2,-4) is the center

Step-by-step explanation:

The standard form of a circle is [tex](x-h)^2+(y-k)^2=r^2[/tex] where (h,k) is the center and r is the radius.

So if you compare

[tex](x-h)^2+(y-k)^2=r^2[/tex]

and

[tex](x-2)^2+(y+4)^2=r^2[/tex],

you should see that

[tex]h=2[/tex]

[tex]-k=4[/tex] which implies [tex]k=-4[/tex]

[tex]r^2=6[/tex] which implies [tex]r=\sqrt{6}[/tex].

So the center is (h,k)=(2,-4).

The radius is [tex]\sqrt{6}[/tex].

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