anonymous
  • anonymous
What is the value of a for the attached circle in general form?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
anonymous
  • anonymous
**no answer choices!! and idk how to solve :(
UnkleRhaukus
  • UnkleRhaukus
factor the equation using prefect squares

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More answers

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

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