Given the following:

x^2+y^2-6y+4x-16=0

how do I find themeasure of the radius?

Printable View

- May 15th 2009, 06:01 PMs3aConics Circle Super simple question
Given the following:

x^2+y^2-6y+4x-16=0

how do I find the**measure of the radius**? - May 15th 2009, 06:40 PMpankaj
$\displaystyle

x^2+y^2-6y+4x-16=0

$

$\displaystyle

(x+2)^2+(y-3)^2=29

$

Obviously, the radiius is $\displaystyle \sqrt{29}$

ALITER:

The general equation of a circle is

$\displaystyle

x^2+y^2+2gx+2fy+c=0

$

the centre being $\displaystyle (-g,-f)$ and the $\displaystyle radius=\sqrt{g^2+f^2-c}$

As per your question $\displaystyle g=2,f=-3,c=-16$

Therefore

$\displaystyle

radius=\sqrt{2^2+(-3)^2-(-16)}=\sqrt{29}

$ - May 16th 2009, 09:22 AMs3a
Can you elaborate on/explain the steps involved between:

x^2+y^2-6y+4x-16=0

and

(x+2)^2+(y-3)^2=29

please? - May 16th 2009, 10:46 AMyeongil
Are you familiar with completing the square?

$\displaystyle x^{2} + y^{2} - 6y + 4x - 16 = 0$

Rearrange terms:

$\displaystyle x^{2} + 4x + y^{2} - 6y = 16$

Complete the square:

$\displaystyle x^{2} + 4x + 4 + y^{2} - 6y + 9 = 16 + 4 + 9$

Factor and simplify:

$\displaystyle (x + 2)^{2} + (y - 3)^{2} = 29$

01 - May 16th 2009, 12:37 PMs3a
In the "Completing the Square" step, how did you make 4 and 9 randomly appear? I understand how you need it on both sides but I don't understand how you "created" them.

- May 16th 2009, 01:22 PMReferos
You can "create" anything in one side of the equation as long as your create the same thing on the other side. So, 4 and 9 were added to the left side of the equation so that the square could be completed, but they had to also be added to the right side to balance things out.

- May 16th 2009, 01:41 PMs3a
But why 4 and 9? Why not other numbers?

- May 16th 2009, 08:16 PMmr fantastic