what is the radius of a circle with the equation x^2+y^2+6x-2y+3=0 ?
soooo heres what i got
Do you know: the standard form of a circle and completing the square?
We'll need both to solve this problem
x^2+6x+....+y^2-2y+....= x^2+6x+ 3^2 +y^2-2y+ (-1)^2 = which i got (x+3)^2 + (y-1)^2= ???? dont know radius
ohh the beginning equation is this lol forgot
ohh ic ,here what you can fix with ur steps x^2+6x+ y^2-2y =-3 x^2+6x+....+y^2-2y+....= -3 + .... + ....
\[x^2+y^2+6x-2y+3=0\]
yep :) then x^2+6x+ 3^2 +y^2-2y+ (-1)^2 =-3+ 3^2+ (-1)^2
Ok now take the numbers that you includes on the left side to complete the square to the other side, because what you do to one sides you have to do to the other
so add 3^2 and (-1)^2 to the right side of the equation
doulikepiecauseidont 's question is very pertinent, and I'dd suggest that you, @meln_n, review both concepts: Do you know: the standard form of a circle and completing the square?
And bring the constant of 3 to the other side so you would subtract it from the number you just added to the right side
sooo finally product \[(x+3)^2+(y-1)^2=7^2\]
^_^ nice work @meln_n
yay soo would radiu be 7 or 49 ???
What do you think? Look carefully at the std. eq'n. of a circle.
i believe 7 because in the equation theres a r^2
remember this formula for circle equation \(\large (x-x_0)^2+(y-y_0)^2=r^2\) such that \( (x_0,y_0) \) is the center and r is the radius
hmm
yes n_n remember
@jdoe0001 why hmming :P so @meln_n whats ur answer ?
heheh
7 n_n ...
why hehehing nw :P
correct so in the std. eq n, the would be square so the radius squared gives you 7 so what is the radius?
x^2+6x+ 3^2 +y^2-2y+ (-1)^2 =-3+ 3^2+ (-1)^2 ^ -3+9+1 = 7 \(\ne 7^2\)
yay ! thank you everyone Conic section is getting to me -_- lol
Here's a graph of the circle just to confirm your answer of the radius http://www.wolframalpha.com/input/?i=x%5E2%2By%5E2%2B6x-2y%2B3%3D0
XD haha lol yeah sry meln
@meln_n check what @jdoe0001 said xD
doulikepiecauseidont I think that deserves a nice apple pie :P
Well, not for me
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