Question

how do you find the radius of x^2+y^2+4x+6y+9=0. what is the working out to find the radius please ?

Answer 1

You have to manipulate it to look like this...
\[\Large (x-h)^2 + (y-k)^2 = r^2\]
Then the radius would be r.

Answer 2

yeah i know but the problem is that i keep making a mistake that i cant find. i end up with the right centre co-ordinates but the radius is wrong could you please do the working out so i can understand

Answer 3

Of course. It all boils down to a nifty little process called "completing the square"
Let's write your equation here...
\[\Large \color{red}{x^2}+\color{blue}{y^2}+ \color{red}{4x}+\color{blue}{6y}+\color{black}9 = 0\]

Answer 4

You'll want to group the x's and the y's together, like so...
\[\Large \color{red}{x^2+4x}+\color{blue}{y^2+6y}+9 = 0\]

Answer 5

Now, to complete the square given the form
\[\Large z^2 + pz\]
You take the coefficient of the term where the variable is raised to the exponent 1, and that's p.
Get half of it, and then square it. That's \(\Large \left(\frac{p}{2}\right)^2\)

You add it to the expression, but ALSO SUBTRACT IT to keep it equal to the original expression....
\[\Large = z^2 + pz \color{green}{+\frac{p^2}{4}}\color{orange}{-\frac{p^2}{4}}\]

Such that it is now equal to...
\[\Large \left(z+\frac{p}2\right)^2 -\frac{p^2}4\]

Answer 6

So, let's complete the square, shall we?
I'll do the x, you do the y.
\[\Large \color{red}{x^2+4x}+\color{blue}{y^2+6y}+9 = 0\]

So, the coefficient of the x (with no exponent) is 4. Half of 4 is 2, and square of 2 is 4.
So we add 4 and we also add 4 to the other side, to balance it out...
\[\Large \color{red}{x^2+4x}\color{orange}{+4}+\color{blue}{y^2+6y}+9 = 0\color{orange}{+4}\]

Answer 7

I cant solve the part where i got upto which is (x+2)^2 + (y+3)^2=4+9+9

Answer 8

Ahh... I see :)
why do you have two "+9"s there?

Answer 9

One of those 9's came from the left side, I'm sure of it.
It was +9 on the left side, so doesn't that mean it becomes -9 when you bring it to the right? ;)

Answer 10

Simplifying...
\[\Large \color{red}{(x+2)^2}+\color{blue}{y^2+6y}+9 = 4\]
Now, completing the square of the y's

\[\Large \color{red}{(x+2)^2}+\color{blue}{y^2+6y}\color{green}{+9}+9 = 4\color{green}{+9}\]

Simplifying again...
\[\Large \color{red}{(x+2)^2}+\color{blue}{(y+3)^2}+9 = 13\]

Answer 11

Still in doubt? :)

Answer 12

thank you very much you've helped me alot :)

Answer 13

No problem :)

Answer 14

\[\Large \color{red}{x^2+4x}+\color{blue}{y^2+6y+9} = 0\]
\[\Large \color{red}{x^2+4x+4}+\color{blue}{y^2+6y+9} = 4\]

Answer 15

the y parts are already a complete square

Answer 16

Me and my formulaic self
:3