Question

Center and Radius of x2 + y2 + 10x – 6y + 18 = 0? :)

Answer 1

It would be useful to have a form more like \((x - h)^2 + (y - k)^2 = r^2\) here.

Are you familiar with completing the square?

Answer 2

Nope, show me how you'd get this answer step by step? :)

Answer 3

try to make factors in term of x and y...by addind and substracting constants..

Answer 4

???

Answer 5

The general idea of completing the square is taking a form like this:
\(x^2 + ax\)
and adding and subtracting something to find a form more like \((x + n)^2 - m\)

The process is a little abstract. In general... \((x + n )^2 = x^2 + 2n x + n^2\). If we have \(x^2 + 2nx\), we add an \(n^2\) to simplify that into \((x + n)^2\), and subtract a \(n^2\) to balance out what we just added on so the value is the same as when we started.

To change \(2n\) into \(n^2\), we divide by two and then square it. That'll be what we do to the linear coefficient in every case of completing the square.

Answer 6

centre would be (-5,3) do u know how?

Answer 7

and radius would be 4

Answer 8

How? :)

Answer 9

For a real example, which may make that ^ seem simpler... :P
We have two quadratics to complete.
\(x^2 + 10x\) and \(y^2 - 6y\)
"To change \(2n\) into \(n^2\), we divide by two and then square it. That'll be what we do to the linear coefficient in every case of completing the square."
For \(x^2 + 10x\), the "2n" is 10. 10/2 = 5. Then we square it, 5^2 = 25.
\(x^2 + 10x + 25 - 25\)
\((x + 5)^2 - 25\)

For \(y^2 - 6y\), its not too different from above, just a small adjustment with the negative. (-6/)/2 = -3, we square it to get (-3)^2 = 9.
\(y^2 - 6y + 9 - 9\)
Minor adjustment is that \(y^2 - 6y + 9\) factors into \((y - 3)^2\) rather than \((y + 3)^2\).
\((y - 3)^2 - 9\)

Answer 10

compare this equation with the general form of the circle... its x^2+y^2+2gx+2fy+c=0
find g and f and c...centre is (-g,-f) and radius is \[\sqrt{g ^{2+f^{2}}-c}\]

Answer 11

now calculate ...its very simple :)

Answer 12

okay... how do you solve that equation? It has so manyy variables....

Answer 13

In the problem...
\(x^2 + y^2 + 10x - 6y + 18 = 0\)
Pair up the quadratics to complete:
\(\color{blue}{x^2 + 10x} + \color{red}{y^2 - 6y} + 18 = 0\)
I already did the completing above, so I could just replace the red/blue. Notice, however, that it is the same value because we added and subtracted the same quantity.
\(\color{blue}{(x + 5)^2 - 25} + \color{red}{(y - 3)^2 - 9} + 18 = 0\)
\(\color{blue}{(x + 5)^2} + \color{red}{(y - 3)^2} - \color{purple}{34} + 18 = 0\)
\(\color{blue}{(x + 5)^2} + \color{red}{(y - 3)^2} - 16 = 0\)
\((x + 5)^2 + (y - 3)^2 = 16\)

Answer 14

@HopelessMathStudent two variables are here just g and f and c is the constant

Answer 15

simple to solve rather to convert first into standard and then see what is the radius and cntre

Answer 16

So its 5,-3 and 16? or would iI need to sqrt 16 and get four?

Answer 17

yeah, sqrt 16 since the general form is r^2. r^2 = 16 (r > 0), so r = 4

Answer 18

and its actually opposite: (-5, 3)

Answer 19

So its center 5,-3 and radius 4 ? :)

Answer 20

how??

Answer 21

in the equation is a positve(the five)

Answer 22

yes.... (-5.3_ and then radius is 4

Answer 23

note that the general form is (x - h)^2, and we have (x + 5)^2.
If we want (x - h)^2, we have (x - -5)^2

Answer 24

geez access denied's work looks like a Error on windows lol... i guess that's becausei spread my work out lol)

Answer 25

or x=-5
x+5=0

Answer 26

hmm, do you not see something like this? http://puu.sh/1yBD7
I sometimes worried if it would mess up for others... :P

Answer 27

that is why x=-5 rather than positive

Answer 28

okay and the 3 is positive?

Answer 29

if acces denied says so then yes

Answer 30

I didn't actually do the problem ahahah

Answer 31

yep
(y - 3)^2, (y - k)^2 is the general form equivalent. k=3 directly. :)

Answer 32

Thank you so much. I really appreciate it!

Answer 33

I would recommend looking into completing the square & practicing it. It can come up in a lot of places, including here. :P

Answer 34

Also, you're welcome! :)

Answer 35

Got it!