Question

Find the center and radius of the circle with the given equation. Then graph the circle. x^2+y^2=16?

Answer 1

Circle with center = (0,0) and radius = r
has the equation of the type: \(x^2 + y^2 = r^2 \)

Answer 2

compare this standard equation with your question.

r^2 = 16
so r = .. ?

Answer 3

r=4?

Answer 4

yes

Answer 5

So do I do x^2+y^2=4

Answer 6

you're asked about center and radius
\(Center \equiv (0,0), r = 4 \)

Answer 7

"So do I do x^2+y^2=4" Be a little more specific: "How do I transform x^2+y^2=4 into standard form and thus find the center and radius?"

Answer 8

How did you find the center?

Answer 9

Notice that both x and y are expressed as squares: x^2, y^2.
This is equivalent to (x-0)^2, (y-0)^2.

Answer 10

circle equation
\(\Large (x-h)^2+ (y-k)^2 =r^2\)

compare it with x^2+y^2 = r^2

h,k = 0,0
r = 4

Answer 11

That's where the center, (0,0), came from.

Answer 12

Borrowing from hartnn, \[\Large (x-h)^2+ (y-k)^2 =r^2\]

Answer 13

contains the info you need about the center of the circle; it is the point (h,k).

Answer 14

(x+3)^2 + (y-9)^2 = r^2 has center (-3,9). That's (h,k) in this situation.

Answer 15

Okay. So the equation is (x-0)^2+(y-0)^2=4?

Answer 16

that's partially correct. Your answer is equivalent to x^2 + y^2 = 2^2.

Actually, you were given an equation with 16 on the right side.

On the right side of this equation, you must type either 4^2 or 16. 4^2 is better.

The radius is r=4, as previously discussed.

Answer 17

Okay so (x-0)^2+(y-0)^2=16?