Question

Question 4: What is the center of the circle with standard form (x-2)2+(y+4)2=16?

Answer 1

Answer:

2,-4

Step-by-step explanation:

I think I did it right but if I didn't you can blame me smh

Answer 2

Answer: second option.

Step-by-step explanation:

The equation of the circle in "Standard form" or "Center-radius form" is the following:

 $(x - h)^2 + (y - k)^2 = r^2$

Where the center of the circle is at the point (h, k) and the radius is "r".

In this case you have the following equation of the circle written in Standard form:

$(x-2)^2+(y+4)^2=16$

You can identify that:

$h=2\\k=-4$

Therefore, the center of the given circle is at this point:

$(2,-4)$