Question

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

Answer 1

Answer:

(2,-4) is the center

Step-by-step explanation:

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

So if you compare

$(x-h)^2+(y-k)^2=r^2$

and

$(x-2)^2+(y+4)^2=r^2$,

you should see that

$h=2$

$-k=4$ which implies $k=-4$

$r^2=6$ which implies $r=\sqrt{6}$.

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

The radius is $\sqrt{6}$.