Question

using the following equation, find the center and radius of the circle.
x^2+2x+y^2+4y=20

medal and fan will be given

Answer 1

@IMStuck hi can you help me please

Answer 2

@ganeshie8 hi can you help me please

Answer 3

@phi can you help me please

Answer 4

@mathmale hey can you help me with this please

Answer 5

@mathstudent55 hi can you help me please

Answer 6

Given the equation x^2+2x+y^2+4y=20, we must put it into the form of the standard equation of a circle with center (h,k) and radius r:\[(x-h)^2+(y-k)^2=r^2\] and so ...

Answer 7

yess

Answer 8

Starting with the given x^2+2x+y^2+4y=20, complete the square twice, first that of x^2+2x and next of y^2+4y.
Have you done "completing the square" before?

Answer 9

no ihavent

Answer 10

I'll guide you thru completing the square for x^2+2x:

The coefficient of 'x" is '2', right?
Take half of this '2'
Square the result.
Do this now, please.

Answer 11

would it be one?

Answer 12

and isn't there a simpler way of doing this

Answer 13

because i did something similar to this in trig

Answer 14

Yes. Half of 2 is 1; the square of 1 is 1. Yes.
Add this to x^2 + 2x, and then subtract the same quantity. Do this now, please.

No...there is no easier way. ;)

Answer 15

lol ok... and 1+2 =3

Answer 16

i was supposed to substitute x for 1 right?

Answer 17

You have x^2 + 2x.
You identify the coefficient of x (which happens to be 2); you halve that, and square the result (obtaining 1), and then you add and immediately subtract that 1 from x^2 + 2x:

x^2 + 2x + 1 - 1. Where did your '2' come from?

Answer 18

gabs?

Answer 19

yes sorry

Answer 20

I'm trying to read it again so I can understand

Answer 21

wait one second I'm getting my notes

Answer 22

Need quicker responses from you, please.

Re write x^2 + 2x as
x^2 + 2x + 1 -1 (which is exactly equivalent).

then your problem becomes

x^2 + 2x + 1 - 1 + y^2 +4y = 20

Answer 23

i can't find them

Answer 24

Please complete the square of y^2 + 4y in exactly the same fashion. Your expression?

\[x^2+2x+1-1+y^2+4y +?-?=20\]

Answer 25

Please finish this.

Answer 26

Hint: what is the coefficient of y?

Answer 27

Gabs? I need to know that you're there and are about to type something. If this is a bad time for you, perhaps we could finish later.

Answer 28

I'm reading sorry

Answer 29

coefficient of y is 4

Answer 30

and half of that is what?

Answer 31

2

Answer 32

so you subtract 2 from the equation right

Answer 33

And the square of 2 is what?

Answer 34

No, you don't subtract 2 from the equation.

Answer 35

the square of 2 is ... ?

Answer 36

4

Answer 37

So, you have y^2 + 4y and must now add 4 to that and then immediately subtract 4. What do you ob tain from following these steps?

Answer 38

x and y

Answer 39

Actually, your y^2 + 4y becomes
y^2 + 4y + 4 - 4, doesn't it? We did the same thing with x^2 + 2x; we got x^2 +2x + 1 - 1, didn't we?

Answer 40

yes and then what do we do?

Answer 41

Rewrite the whole equation:

x^2 + 2x + 1 - 1 + y^2 + (you finish it, please)

Answer 42

4x-4

Answer 43

Don't you mean 4y + 4 - 4? leading to

x^2 + 2x + 1 - 1 + y^2 + +4y + 4 - 4 = 20 ?

Answer 44

Lot of detail here, but it's essential that you undrstand these procedures.

Answer 45

Note, please, that x^2 + 2x + 1 is the square of (x+1)^2, so we can re-write

x^2 + 2x + 1 - 1 + y^2 + +4y + 4 - 4 = 20 as

(x+2)^2 -1 + y^2 + +4y + 4 - 4 = 20

Answer 46

Can you finish this? Rewrite y^2 + 4y + 4 as what perfect square?

Answer 47

ok .... how is is this going to find the radius

Answer 48

ill finish this thanks

Answer 49

for your patience

Answer 50

You should end up with

(x+1)^2 + (y+2)^2 -1 -2 = 20 -1 -2 = 17.

Compare this to \[(x-h)^2+(y-k)^2=r^2\] Thru this comparison, can you identify the value of 'h'?

Answer 51

1 ?

Answer 52

Actually, it'd be -1.

Here's why: (x-h)^2 = (x+1)^2
So -h=+1, or +h=-1. OK?

Answer 53

ok thank you

Answer 54

:)

Answer 55

You're welcome, gabs! Good luck!

Answer 56

@phi can you help me please i didn't understand with the other person

Answer 57

@thomaster hey can you help me

Answer 58

how far did you get ?

Answer 59

i honestly didn't understand anything he said

Answer 60

i just need to find the radius and center... and did he get it right so far ? @phi

Answer 61

I am sure he did.

To do this, you use something called
"completing the square"

ring any bells.?

Answer 62

no i was never taught this

Answer 63

actually i understand so far but what about the rest? @phi

Answer 64

start with
x^2+2x+y^2+4y=20

add 1 to both sides
x^2 + 2x + 1 + y^2 +4y = 20+1
add 4 to both sides

(x^2 + 2x + 1) + (y^2 +4y +4) = 20+1+4

by algebraic magic we have (x^2 + 2x + 1) = (x+1)^2
and (y^2 +4y +4) = (y+2)^2
and on the right side 20+1+4= 25

we get
(x+1)^2 + (y+2)^2 = 5^2

Answer 65

and the radius would be 5^2

Answer 66

and the center 1,2 right

Answer 67

we want to match that up with
\[ (x-h)^2 + (y-h)^2 = r^2\]

we re-write x+1 as x- (-1) and y+2 as y - (-2)
\[ (x - (-1))^2 + (y - (-2))^2 = 5^2 \]

Answer 68

you have the right idea, but the details matter.

5^2 "matches up" with r^2
so r is not 5^2. r is 5

Answer 69

and the center is -1,-2

Answer 70

thank you soooo much you explained it much better :)