iheartfood
What is the value of a for the attached circle in general form?
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iheartfood
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iheartfood
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**no answer choices!!
and idk how to solve :(
UnkleRhaukus
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factor the equation using prefect squares
iheartfood
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which equation??
iheartfood
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like the one of the circle?? (which i find??) or the general form?
UnkleRhaukus
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\[x^2+ax+y^2+by+c=0\]
\[\left(x+\frac a2\right)^2-a^2+y^2+by+c=0\]
iheartfood
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why whoops?? haha :P am i not seeing something ur seeing??
UnkleRhaukus
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(i forgot to square ) (fixed now)
iheartfood
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ohhh okay i see :P so what do i do now?
UnkleRhaukus
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factor the y terms using the same technique , then move the three constants outside the brackets to the right hand side
iheartfood
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(x+a/x)^2=(-a+y+by)^2
?? did i do that right?? :/ idk...
iheartfood
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oops i forgot the +c...
so like this??
(x+a/x)^2=(-a+y+by)^2+c
??? idk :(
UnkleRhaukus
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\[\left(x+\frac a2\right)^2-a^2+y^2+by+c=0\]
\[\left(x+\frac a2\right)^2-a^2+\left(y+\frac b2\right)^2-b^2+c=0\]
\[\left(x+\frac a2\right)^2+\left(y+\frac b2\right)^2=a^2+b^2-c\]
iheartfood
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oh okay :P
but how do i solve for a?
UnkleRhaukus
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compair to the general form of a circle
iheartfood
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|dw:1354600365624:dw|
iheartfood
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or the one in my question?
UnkleRhaukus
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|dw:1354600399190:dw|
iheartfood
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oh sorry! i meant - haha :P so do i refer to that one?? or the one in my problem? this one?|dw:1354600447714:dw|
UnkleRhaukus
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|dw:1354600492668:dw|
UnkleRhaukus
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now look at the graph
iheartfood
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okay
so (h,k) center=(-4,2) and radius=2 ???
iheartfood
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is that right??
but how do i get the value of a from that?
UnkleRhaukus
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\[\left(x+\frac a2\right)^2+\left(y+\frac b2\right)^2=a^2+b^2-c\]
\[\left(x-h\right)^2+\left(y-k\right)^2=r^2\]
iheartfood
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so a= -4/2 ???
iheartfood
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so a=-2 ??
is that right?? my answer is -2??
jim_thompson5910
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center=(-4,2) ---> h = -4, k = 2
radius=2 ---> r = 2
(x-h)^2 + (y-k)^2 = r^2
(x-(-4))^2 + (y-2)^2 = 2^2
(x+4)^2 + (y-2)^2 = 4
Now get this equation into x^2 + y^2 + ax + by + c = 0 form
iheartfood
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okay, lemme give this a shot hahaha idk how close ill get tho!!!
x^2 + y^2 + 4^2 - 2^2 - 4 = 0 ???
idk haha :P
iheartfood
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is that right?? or even close?? :/
jim_thompson5910
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no that's not correct
jim_thompson5910
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(x-h)^2 + (y-k)^2 = r^2
(x-(-4))^2 + (y-2)^2 = 2^2
(x+4)^2 + (y-2)^2 = 4
x^2 + 8x + 16 + y^2 - 4y + 16 = 4
jim_thompson5910
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keep going
iheartfood
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aww darn :(
okay so i simplify that right?
iheartfood
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kk so i get this?
x^2 + 8x + 16 + y^2 - 4y + 16 = 4
= x^2 + y^2 +8x -4y +12 = 0 ??
jim_thompson5910
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not quite
jim_thompson5910
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16+16-4 isn't 12
iheartfood
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ohhh theres another 16!! i didn't see that! my bad!!!
so i get this??
x^2 + y^2 +8x -4y + 28 = 0
iheartfood
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kk so a=8 then??
my answer is 8? :O
iheartfood
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awesome :) thanks so much for all of ur help @UnkleRhaukus and @jim_thompson5910 :)