Question

find the radius of the circle defined by the equation;
3x^2+3y^+42y+186=0 who can help?

Answer 1

*3y^2

Answer 2

so?

Answer 3

dividing with 3 we get
x^2+y^2+14y+62=0
hence we can write it as
(x-0)^2 +(y+7)^2 +13=0
comparing with standard form
answer is sqrt of 13

Answer 4

This needs a nifty trick known as "Completing the square"

Answer 5

But yeah, before that, better make sure that the x^2 and y^2 only have coefficients of 1.

Answer 6

Wait a minute, there's something wrong with this

(x-0)^2 +(y+7)^2 +13=0

Answer 7

I don't think this has any solutions at all...
no circle to speak of

Answer 8

It gives you
\[x^2+(y+7)^2=-13\]

Answer 9

@PeterPan add the 49 to both sides.

Answer 10

It's been done, @Mertsj
That's why 62 became 13

Answer 11

So there is either a typo or no such circle exists.

Answer 12

\[x^2+y^2+14y+62=0\]\[x^2+y^2+14y\color{red}{+49}+62=0\color{red}{+49}\]\[x^2+(y+7)^2+62=49\]\[x^2+(y+7)^2+62\color{red}{-62}=49\color{red}{-62}\]\[x^2+(y+7)^2=\color{green}{-13}\]

Answer 13

Must be a typo @sellweezy96
Please recheck :)

Answer 14

3x^2+3y^2+42y+186=0

Answer 15

Could it possibly be -186?

Answer 16

3x^2+3y^2+42x+42y+186=0 @Mertsj

Answer 17

Okay, so there's a 42x after all :D

Answer 18

Well that's a whole different thing.

Answer 19

First, as @irshadno1 did
so shall you do ^.^
Divide everything by 3. We don't want coefficients of x^2 which are not 1.

Answer 20

x^2+Y^2+14x+14y+62=0 ?

Answer 21

Good.
Now complete the square.
x^2 + 14x + y^2 + 14y + 62

Answer 22

divide it by which

Answer 23

Here's an example of completing the square
\[\large z^2+6z \]
See that 6z? Its coefficient is 6. Get half of that, and then square it.
Half of 6 is 3, and the square of 3 is 9.
Add that.
\[\large z^2+6z+9\color{red}{-9}\]
\[\large (z+3)^2\color{red}{-9}\]

Answer 24

So, back to your circle
\[\large \color{blue}{x^2+14x}+\color{green}{y^2+14y}+62=0\]
You want to complete the square of the blue (x) part?

Answer 25

=(x+7)+49 ?

Answer 26

Almost...
14x --> what's half of 14?

Answer 27

7

Answer 28

and the square of 7?

Answer 29

49

Answer 30

Okay, good :)
So you either add and subtract 49 to the left side
OR you add 49 on BOTH sides... let's go the other way...
\[\large \color{blue}{x^2+14x}\color{red}{+49}+\color{green}{y^2+14y}+62=0\color{red}{+49}\]\[\large \color{violet}{(x+7)^2}+\color{green}{y^2+14y}+62=\color{red}{49}\]

Answer 31

Now do the same to the green (y) part.