Question

How do I find the center and radius of a circle for the equation x^2 +(y-1)^2=4

Answer 1

do you know the general form for the equation of a circle?

Answer 2

if not, this might help: http://www.mathsisfun.com/algebra/circle-equations.html

Answer 3

thank you again :-)

Answer 4

yw :)

Answer 5

It helped but I am still confused again ): I see the formulas but I am not sure which one to use or how to apply the equation to the formulas..

Answer 6

ok, if a circle has its center at (a, b), then what would its equation be?

Answer 7

maybe it would help if you rewrote your equation this way:
\(\large (x-0)^2 +(y-1)^2=4 \)

Answer 8

and if it had a radius r

Answer 9

\[(x-a)^2 + (y-b)^2=r^2\]

Answer 10

?

Answer 11

good - so now try and rewrite your equation in this form

Answer 12

\[x^2 +y^2 -2y-3=0 \]

Answer 13

I ended up with this...

Answer 14

no - do not expand the terms

Answer 15

your original equation is:\[x^2 +(y-1)^2=4\]so the term in y is already in the "right" form (i.e. \((y-b)^2\))

Answer 16

now think of how you can rewrite \(x^2\) as \((x-a)^2\) - what would a=?

Answer 17

4?

Answer 18

so you are saying that \(x^2\) is the same as \((x-4)^2\)?

Answer 19

no, i was moving the 4 over and somehow making that work. i am not sure how i can rewrite it.

Answer 20

maybe this will help:\[x^2=(x)^2=(x-?)^2\]what should the ? be replaced with to make this correct?

Answer 21

1?

Answer 22

ohhh 0?

Answer 23

which means you are saying that:\[x^2=(x-1)^2\]try again...

Answer 24

bingo! well done!

Answer 25

lol it takes me awhile to catch on, thank you for being patient with me.

Answer 26

so now we have:\[(x-0)^2+(y-1)^2=4\]but we need it in the form:\[(x-a)^2+(y-b)^2=r^2\]so think about how you can express 4 as something squared?

Answer 27

two squared, correct?

Answer 28

perfect!
so we ended up with:\[(x-0^2+(y-1)^2=2^2\]so you should now be able to tell me where the center of this circle is and what its radius is

Answer 29

sorry - missed a bracket on the x terms there

Answer 30

the center is (0,1) and the radius is 2?

Answer 31

perfect!

Answer 32

thank you :-)

Answer 33

yw :)

Answer 34

I also had to find the intercepts for that equation. I graphed the circle and found 4 intercepts, but two of them were wrong. My answer was (2,0), (0,3),(-2,0), and (0,-1). The correct answer is \[(-\sqrt{3},0),(\sqrt{3},0), (0,-1), (0,3)\]

Answer 35

what did I do wrong?

Answer 36

you do not need to graph this to find the intercept.
use the same method as I showed you before to find the intercepts - i.e. set x=0 to find the y-intercepts and set y=0 to find the x-intercepts.

Answer 37

oops lol thanks

Answer 38

BTW: I guess since you are fairly new to this site you should know that the usual workflow is:

1. You post a question.
2. One or more people come to your question to help you out.
3. You select the one that best answered your question by clicking on the blue Best Response button next to any one of their replies.
4. You close the question by clicking on the Close button

then the whole process starts again.

Answer 39

I'm sorry

Answer 40

np :)