Question

Need help with finding equation of given circle
center (3, 3)
radius of 2

How do I solve this?

Answer 1

take a look here http://www.purplemath.com/modules/sqrcircle.htm

Answer 2

you already has the center and the radius, you have to do steps backwards

Answer 3

can you walk me through this

Answer 4

one sec

Answer 5

i dont know which format to even use

Answer 6

formula for a circle with center at (h,k) and radius r is

(x-h)^2 + (y-k)^2 = r^2

plug in the numbers...

Answer 7

thx

Answer 8

you have center of (3,3) = (h,k) so h = 3, k = 3
radius = 2 so r = 2

Answer 9

so you can use both of those formula's?

Answer 10

you can expand the quation to get ax^2+by^2+cx+dy=e

Answer 11

is that what you did?

Answer 12

it depends on what information you're given

Answer 13

normally, I'd leave it in the (x-h)^2 + (y-k)^2 = r^2 form unless specifically instructed to expand it. You can easily see the details of the circle in the former form, which is useful. The latter form you can't see diddly unless you do the algebra to convert it back.

Answer 14

yeah, I agree

Answer 15

ok so (x-h)^2+(y-k)^2 after I plug in those numbers what do i do? I mean it'll be just like x-3+y+3=2

Answer 16

and I can't do anything with x-3? can I except combine them

Answer 17

no, it will be (x-3)^2 + (y-3)^2 = 2^2

Answer 18

Yeah I was just using for example

Answer 19

ok nvm

Answer 20

but you didn't have any exponents in there

Answer 21

you have to have the x term squared, and the y term squared, or it won't be a circle.

Answer 22

Im using (x-h)^2+(y-k)^2 right?

Answer 23

yes. notice that if the center of the circle is at (0,0) that equation reduces to
(x-0)^2 + (y-0)^2 = r^2
x^2 + y^2 = r^2
which is just the Pythagorean theorem in disguise!

Answer 24

ok hold on let me do this and see if I understand give me a min

Answer 25

What do I plug in for k? 3?

Answer 26

if the draw tool worked, I'd draw you a picture but the equation for the circle is just showing you all the points that are the same distance away from the center. the distance formula is also the Pythagorean theorem in disguise:

d = sqrt((x2-x1)^2 + (y2-y1)^2)
d^2 = (x2-x1)^2 + (y2-y1)^2

(formula for distance between (x1, y1) and (x2, y2))

Answer 27

our center is at (3,3), right? so the formula expects the center at (h,k) so h = 3, k = 3

Answer 28

ah

Answer 29

Yes if the draw tool was working I'd understand much better

Answer 30

Im more of a visual learner

Answer 31

let me plug it in

Answer 32

it unnecessarily confuses things to use the same number, in my opinion. I would have written a problem where the numbers were all different, just to make it that much clearer which bit goes where.

Answer 33

x-3^2+y-3^2=2^2 correct?

Answer 34

>>x-6+y-6=4

Answer 35

after simplifying it

Answer 36

well centre is 3,3 and radius is 2. equations of a circle is x^2 + y^2 = r^2

Answer 37

no! you MUST have the exponents and parentheses!

(x-3)^2 + (y-3)^2 = 2^2

Answer 38

oh ok

Answer 39

Yeah I know

Answer 40

(x-6)+(y-6)=4

Answer 41

(x-3)^2 + (y-3)^2 = 2^2
(x^2 -3x - 3x + 9) + (y^2 -3y -3y + 9) = 4
x^2 - 6x + 9 + y^2 - 6y +9 = 4
x^2 - 6x + y^2 - 6y + 18 = 4
x^2 - 6x + y^2 - 6y = -14

sorry @dg98 the "NO" was directed at Mr. Ick

Answer 42

is there no way You can draw this out?

Answer 43

@Mr.Ick stop skipping the ^2 !!!

Answer 44

hold on

Answer 45

Sorry, I have to leave, but it's all there. Just read through my posts again.

Answer 46

thats not right

Answer 47

what's not right?

the equation of a circle with center at (3,3) and radius 2 is
(x-3)^2 + (y-3)^2 = 2^2

or

(x-3)^2 + (y-3)^2 = 4

if you expand it, you get

x^2 + y^2 - 6x - 6y + 18 = 4

or

x^2 + y^2 - 6x - 6y + 14 = 0

here's a picture: http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=circle%20with%20center%20(3%2C3)%20and%20radius%202