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iheartfood
Group Title
What is the value of a for the attached circle in general form?
 one year ago
 one year ago
iheartfood Group Title
What is the value of a for the attached circle in general form?
 one year ago
 one year ago

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iheartfood Group TitleBest ResponseYou've already chosen the best response.1
**no answer choices!! and idk how to solve :(
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
factor the equation using prefect squares
 one year ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.1
which equation??
 one year ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.1
like the one of the circle?? (which i find??) or the general form?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
\[x^2+ax+y^2+by+c=0\] \[\left(x+\frac a2\right)^2a^2+y^2+by+c=0\]
 one year ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.1
why whoops?? haha :P am i not seeing something ur seeing??
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
(i forgot to square ) (fixed now)
 one year ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.1
ohhh okay i see :P so what do i do now?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
factor the y terms using the same technique , then move the three constants outside the brackets to the right hand side
 one year ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.1
(x+a/x)^2=(a+y+by)^2 ?? did i do that right?? :/ idk...
 one year ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.1
oops i forgot the +c... so like this?? (x+a/x)^2=(a+y+by)^2+c ??? idk :(
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
\[\left(x+\frac a2\right)^2a^2+y^2+by+c=0\] \[\left(x+\frac a2\right)^2a^2+\left(y+\frac b2\right)^2b^2+c=0\] \[\left(x+\frac a2\right)^2+\left(y+\frac b2\right)^2=a^2+b^2c\]
 one year ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.1
oh okay :P but how do i solve for a?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
compair to the general form of a circle
 one year ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.1
dw:1354600365624:dw
 one year ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.1
or the one in my question?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
dw:1354600399190:dw
 one year ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.1
oh sorry! i meant  haha :P so do i refer to that one?? or the one in my problem? this one?dw:1354600447714:dw
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
dw:1354600492668:dw
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
now look at the graph
 one year ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.1
okay so (h,k) center=(4,2) and radius=2 ???
 one year ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.1
is that right?? but how do i get the value of a from that?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
\[\left(x+\frac a2\right)^2+\left(y+\frac b2\right)^2=a^2+b^2c\] \[\left(xh\right)^2+\left(yk\right)^2=r^2\]
 one year ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.1
so a= 4/2 ???
 one year ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.1
so a=2 ?? is that right?? my answer is 2??
 one year ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.0
center=(4,2) > h = 4, k = 2 radius=2 > r = 2 (xh)^2 + (yk)^2 = r^2 (x(4))^2 + (y2)^2 = 2^2 (x+4)^2 + (y2)^2 = 4 Now get this equation into x^2 + y^2 + ax + by + c = 0 form
 one year ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.1
is that right?? or even close?? :/
 one year ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.0
no that's not correct
 one year ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.0
(xh)^2 + (yk)^2 = r^2 (x(4))^2 + (y2)^2 = 2^2 (x+4)^2 + (y2)^2 = 4 x^2 + 8x + 16 + y^2  4y + 16 = 4
 one year ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.0
keep going
 one year ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.1
aww darn :( okay so i simplify that right?
 one year ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.0
yes
 one year ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.1
kk so i get this? x^2 + 8x + 16 + y^2  4y + 16 = 4 = x^2 + y^2 +8x 4y +12 = 0 ??
 one year ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.0
not quite
 one year ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.0
16+164 isn't 12
 one year ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.1
ohhh theres another 16!! i didn't see that! my bad!!! so i get this?? x^2 + y^2 +8x 4y + 28 = 0
 one year ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.0
yep
 one year ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.1
kk so a=8 then?? my answer is 8? :O
 one year ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.0
yes
 one year ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.1
awesome :) thanks so much for all of ur help @UnkleRhaukus and @jim_thompson5910 :)
 one year ago
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