Ask your own question, for FREE!
Mathematics 22 Online
OpenStudy (anonymous):

What is the equation of a circle whose diameter is 6 cm?

Directrix (directrix):

Are you given the coordinates of the center of the circle?

OpenStudy (anonymous):

No it just says that. Im assuming just to use (0,0) but I dont know because thats all that the question is.

Directrix (directrix):

Look at the attachment I posted above. (h,k) is standard for the coordinates of the center of a "mystery" circle. Note that the formula gives the radius, not the diameter so you will have to adjust the 6.

OpenStudy (anonymous):

x^2+y^2=3

Directrix (directrix):

Why would you assume this: Im assuming just to use (0,0)

OpenStudy (anonymous):

This would be wrong because i dont have the radius?

Directrix (directrix):

>>>x^2+y^2=3 Wrong even if the center were (0,0) @chetan552

Directrix (directrix):

@shelbyv If you know the diameter, how do you get the radius of the circle?

OpenStudy (anonymous):

So @Directrix what is the correct answer?

Directrix (directrix):

Are you saying that there is no diagram? Just an out of the blue question: What is the equation of a circle whose diameter is 6 cm?

OpenStudy (anonymous):

Radius is half of the diameter and yes that is EXACTLY what im sayying.

OpenStudy (solomonzelman):

\[(x-h)^2+(y-k)^2=9\]the center is not specified.

OpenStudy (solomonzelman):

if the center is (0,0)\[x^2+y^2=9\]

Directrix (directrix):

@shelbyv Then, find the equation of a circle with center (h,k) and radius 3. (x - h ) ^2 + (y - k) ^2 = 3^2 is what I would put. Do you have options?

OpenStudy (anonymous):

nope no options at all. Just that question typed on a blank sheet of paper.

Directrix (directrix):

I would put that this and move on: (x - h ) ^2 + (y - k) ^2 = 3^2

OpenStudy (anonymous):

Okay. thanks so much.

Directrix (directrix):

If that is said to be wrong, then defend it. You have all the "defense" in this thread.

OpenStudy (anonymous):

Okay :)

OpenStudy (anonymous):

Ya i said same radious=diameter/2 @Directrix

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!