Question

Find the center and radius then graph each equation. (x-3)^2 +(y+4)^2 -9=0

Answer 1

you want the quick answer, or the way to know how this works?

Answer 2

well i only have 2 so it's up to you. i don't mind learning if you explain it well

Answer 3

Well, let's take the simplest case, a circle centered on the origin (0,0).

a circle is the set of all the points that are an equal distance (the radius) from its center.

Answer 4

yah

Answer 5

Now, let's draw a little picture:

Answer 6

oops, didn't label the radius, but you know where it is. that's a little right triangle, just like we love with the Pythagorean theorem. the radius/hypotenuse is going to satisfy the equation x^2 + y^2 = r^2, and the circle is just all the points crossed over by the tip of the triangle as we sweep through all possible values of x and y

Answer 7

with me so far?

Answer 8

oh ok i didnt know we would use that therom but it makes sense

Answer 9

yep. now, what if we wanted a circle with a center somewhere else? first, let's say we have a radius of 1:

Answer 10

ok yah so that is a mini circle

Answer 11

we can test out our equation and make sure that it goes through those points. let's try (1,0):

1^2 + 0^2 = 1^2, that works, let's try (0,1):
0^2+1^2=1^2 that works and we can see the other 2 will as well.

Answer 12

now, if we move the circle 1 unit to the right, those four points (1,0), (0,1), (-1,0),(0,-1) become (2,0),(1,1),(0,0), (1,-1), right?

Answer 13

yah that's right

Answer 14

question is, what do we do to the circle's equation to make that happen?

Answer 15

sorry to intrude but.. is the center at (3,-4)?

Answer 16

He is giving me an example

Answer 17

well, if we subtract 1 from x before plugging it in, let's see what happens:

\[(x-1)^2+y^2= 1^2\]
We'll try the point (2,0) which we think should be the East compass point:
\[(2-1)^2+0^2 = 1^2\]\[1^2+0^2=1^2\checkmark\]Let's try the North compass point:\[(1-1)^2+1^2=1^2\checkmark\]Now the South compass point\[(1-1)^2+(-1)^2=1^2\checkmark\]and you can see the West one works as well

Answer 18

yes yes...

Answer 19

So, to move the circle along the x-axis, we either add or subtract the shift from the value of x before squaring. And because the formula is symmetrical, I trust you can see the same will be true for moving up and down the y-axis, if we subtract/add from the y value,right?

Answer 20

1 sqared unit

Answer 21

so that means we can write the formula for the circle with radius 1 and center at (h,k) as \[(x-h)^2+(y-k)^2=1\]

Answer 22

but i don't get what this has to do with my problem -_-

Answer 23

Now for the radius. Let's go back to our center at the origin just to make life easier:
say we have radius 2, then the East point is at (2,0)
\[(2-0)^2+(y-0)^2 =4\]But our radius is 2, which is the square root of 4, so that means the right hand side is the radius squared. Let's do the same with radius 3. East compass point at 3,0, \[(3-0)^2+(0-0)^2=9\] and 9=3^2 so clearly the right hand side of the equation is \(r^2\).

Answer 24

SO, our formula for any circle having center \((h,k)\) and radius \(r\) is
\[(x-h)^2+(y-k)^2=r^2\]

Compare that with your problem statement, and read off the answers!

Answer 25

You just need to rearrange your original equation slightly to match by adding 9 to both sides...

Answer 26

having trouble rearranging

Answer 27

understanding the translations we did here to move the center around the coordinate plane will be handy when you start doing ellipses, parabolas, hyperbolas, etc. general rule of thumb: to move the figure to the right, subtract from x before feeding it into the function. to move it to the left, add to x before feeding it into the function. to move the figure up, add to the output of the function, and to move it down, subtract from the output of the function. multiplying the result of the function scales the vertical range, multiplying the input of the function scales the horizontal.

Answer 28

\[(x-3)^2 +(y+4)^2 -9=0\]Add 9 to both sides
\[(x-3)^2+(y+4)^2-9+9=0+9\]\[(x-3)^2+(y+4)^2=9\]
Compare with \[(x-h)^2+(y-k)^2=r^2\]

Answer 29

well h= 3 and k=4 and r=3

Answer 30

not so fast...look carefully at the y term...

Answer 31

oh

Answer 32

its negative

Answer 33

-k = +4
k = -4

Answer 34

so where is the center of the circle?

Answer 35

it's at (h,k), right?

Answer 36

idk why?

Answer 37

didn't we just work all that out? that formula that I derived and lined up below the formula for your circle

Answer 38

oh
i was confused on what h and k and r stand for

Answer 39

If you have a circle with center at \((h,k)\) and radius \(r\), the formula is \[(x-h)^2+(y-k)^2=r^2\]
Our circle is \[(x-3)^2+(y+4)^2=9\] or \[(x-3)^2+(y-(-4))^2=3^2\]
What is the center and radius of our circle?

Answer 40

well um (3,-4) and i am not sure about the radius :(

Answer 41

good on the center. if the right hand side of the equation represents \( r^2\), and it equals 9, what's the problem with finding the radius, \(r\)?

Answer 42

3