circle

unique · 2006-10-30 02:51:18

Find the equation of teh circle whose center is at (3,2)
and whose radius is 5.
the answer has ot be in standard form.
can anyone tell me how to do this

luca-deltodesco · 2006-10-30 03:44:18

the equatin for a circle is

for yours this is

now, this is for a circle with x = y = 0

to translate a graph c units to the right, you have to take c away from x, similarly for y, so for you have

im not sure what it means by standard form, whether it wants it in that for, or in expanded form by

unique · 2006-10-30 03:54:04

umm thanks
i guess they are asking for that....
standard form is > x^2 + y^2 = r^2
so i think
it has to be
(x - 3)^2 + (y - 2)^2 = 5^2

and
x^2 + y^2 - 8 = 0 is equal to general form.
THANKS = )

mathsyperson · 2006-10-30 06:34:31

I'd say that (x-3)² + (y-2)² = 5² was the standard form.

If you did expand it out though, you'd get x² - 6x + y² - 4y - 12 = 0.

luca-deltodesco · 2006-10-30 08:37:25

mathsyperson wrote:

I'd say that (x-3)² + (y-2)² = 5² was the standard form.
If you did expand it out though, you'd get x² - 6x + y² - 4y - 12 = 0.

yeh i know, i dont know where i got that thing from but i was busy eating at the time, nevermind

Math Is Fun Forum

#1 2006-10-30 02:51:18

circle

#2 2006-10-30 03:44:18

Re: circle

#3 2006-10-30 03:54:04

Re: circle

#4 2006-10-30 06:34:31

Re: circle

#5 2006-10-30 08:37:25

Re: circle

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