ok so I'm feeling quite dumb today, and can't figure this out.

Quote :

" Find the standard form of the equation of the circle having the following properties: center at the origin containing point (5,-2) "

I'm am completely lost with the "center at the origin" part. I just simply do not know what to do here.
lil help tdub?

10/15/2007 6:46:05 AM

what is the standard form for the equation of a circle with center (a,b) and radius r?

1st answer that.

what is the "origin" on the coordinate system? what point does that refer to? what are its coordinates?

then answer that.

so now you know what the center of the circle is [i.e., the (a,b)]. now you need the radius. how will you find the radius using the center (which you know now) and the point (5,-2) which they tell us is on the circle? think about what radius of a circle means. what distance is called the radius?

then answer that.

now that you have the center and the radius, you can simply plug them into the standard form of the equation of a circle.

10/15/2007 7:04:40 AM

x^2+y^2=r^2 standard form

origin, have no damn clue, unless its 0,0

am I thinking right so far?

10/15/2007 7:10:42 AM

the origin is (0,0) like you said, and if the circle contains point (5,-2), then that point lies (lay/lie?), on the edge of the circle. You can determine the radius by finding the distance between the origin (0,0) and a point on the edge (5,-2), and then proceed to the standard form for the equation of a circle.

10/15/2007 7:53:35 AM

damn dawg, where you been? origin is from middle school, yo. anyway, so origin is (0,0) as you said.

standard form of circle equation is (x-a)^2 + (y-b)^2 = r^2, not what you said.

where (a,b) is the center of the circle, and r is the radius.

but if center of circle (a,b) = origin = (0,0), then the equation of YOUR circle is reduced to:

(x-0)^2 + (y-0)^2 = r^2, which simplifies to what you wrote.

but you still need radius. and remember, you have to put r^2 in the equation, not r.



p.s. so x^2 + y^2 = r^2 is basically the standard form of the equation of a circle centered at the origin. but circles don't have to be centered at the origin... just that yours is. if not centered at the origin, then the equation is with the a and b in it, as i gave above.


[Edited on October 15, 2007 at 8:02 AM. Reason : ]

10/15/2007 8:00:35 AM

for a circle centered at (h,k), the equation is (x-h)^2+(y-k)^2=r^2

so:
(x-5)^2+(y+2)^2 = (radius)^2

10/15/2007 8:14:36 AM

Quote :

"(x-5)^2+(y+2)^2 = (radius)^2"

WRONG.

Circle is NOT centered at (5,-2).

Read 1st post again.

10/15/2007 8:18:48 AM

Quote :

"damn dawg, where you been? origin is from middle school, yo. anyway, so origin is (0,0) as you said. standard form of circle equation is (x-a)^2 + (y-b)^2 = r^2, not what you said. where (a,b) is the center of the circle, and r is the radius. but if center of circle (a,b) = origin = (0,0), then the equation of YOUR circle is reduced to: (x-0)^2 + (y-0)^2 = r^2, which simplifies to what you wrote. but you still need radius. and remember, you have to put r^2 in the equation, not r. p.s. so x^2 + y^2 = r^2 is basically the standard form of the equation of a circle centered at the origin. but circles don't have to be centered at the origin... just that yours is. if not centered at the origin, then the equation is with the a and b in it, as i gave above"

ok, its comin to me, since the center is (0,0) and a point is (5,-2); I solve for the radius now. ok.
yeah that whole "origin" thing threw me off.

first math class since, umm, well shit, since high school. just about 10 years ago, so I admit I'm a lot rusty. doin pretty good but occasionally I'll come across something that'll make me go "wft? how do you that?"


many thanks

10/15/2007 8:27:36 AM

Quote :

"just about 10 years ago"

sorry didn't know that! took a break after high school till college?

anyway, you are welcome.

you know how to find the radius, right? using the distance formula.

10/15/2007 8:37:21 AM

yeah I"ll all over the radius, done it a few times since I posted. that center of origin is what threw me off.

yeah ten years since "real math" lol.
military, several years deployed. few years on campus but managing to avoid math like the plague/vd.
now that I'm still deployed and in my senior year(s), I gotta get it done and few others so I can finally finish up. been working on my degree since '01

but yeah, I'll probably pop back in here to get a lil help with the alge-don't-tase-me-bra so I'm really glad ya'll got it locked. interestingly enough, I was a comp sci major for a few years and still managed to avoid math lol.

10/15/2007 8:43:40 AM

Quote :

"WRONG. Circle is NOT centered at (5,-2)."

yeah, I'm not awake yet, sorry.

10/15/2007 8:45:14 AM

oh damnit, failed that test. got the circle questions somewhare right.

damnit.

10/15/2007 10:18:53 PM

Here is another circle centered at origin with (5,-2) on it,

abs(x)+abs(y) = 7

circle becomes the square, well diamond actually.

See it isn't hard at all to imagine a square circle, it just depends on your idea of distance.

10/15/2007 10:20:32 PM

^Are you using Taxicab distance instead of Euclidean?

10/16/2007 7:58:15 AM

^^Right, that's what a circle looks like to manhattanites.

[Edited on October 16, 2007 at 8:01 AM. Reason : ^of course]

10/16/2007 8:00:37 AM

^^yes
^ ah ha.

10/16/2007 5:06:51 PM