Question

Find the center and radius of the circle with equation x^2 + y^2 - 6x + 10y + 9 = 0

Answer 1

if given a circle equation : x^2+y^2+AX+By+C=0, then
the centre of circle at point : (a,b)
with a = -A/2 and b = -B/2
for radius (r) = sqrt((1/4)A^2 + (1/4)B^2 - C)

Answer 2

^?

Answer 3

I like the explanation on this page:
http://www.analyzemath.com/Calculators/Cen_Rad_Cal.html

Answer 4

...algebra.....scary......

Answer 5

too many variables

Answer 6

A=-6, B=10, C=9

Answer 7

Basically, you can look at the \(x^2 + ax\) part of the equation and use completing the square to rewrite it.
Then you can look at the \(y^2 +by\) part of the equation and use completing the square to rewrite that.
If you don't love completing the square, just use the formula given. Gather all of the constant terms on one side of the equation, and suddenly the equation is in standard form.

Answer 8

completing squares huh..

Answer 9

\(x^2 + a x = (x + a/2)^2 - (a/2)^2 \)
For your problem, a=-6

Answer 10

hmm

Answer 11

How familiar are you with completing the square? I'm sure you've seen it come up.

Answer 12

not that good

Answer 13

i mean, A,B, and C are constanta

Answer 14

i'll try my best though

\[x^2 - 6x + y^2 + 10y = -9\]
\[(x - 3)^2 + (y+ 5)^2 = -9 + 9 + 25\]

Answer 15

so then it becomes \[(x-3)^2 + (y+5)^2 = 25\]

Answer 16

it will work also

Answer 17

so then the center is (3,-5) and radius is 5...

Answer 18

yes, u are right

Answer 19

feel free to call me wrong though

Answer 20

i suppose that was a no

Answer 21

Correct =)

Answer 22

Completing the square is an interesting concept that is actually fairly easy to wrap your head around if you see it done geometrically.

Answer 23

i'm not really used to doing completing square method