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Mathematics 18 Online
OpenStudy (anonymous):

What is the value of a for the attached circle in general form?

OpenStudy (anonymous):

OpenStudy (anonymous):

**no answer choices!! and idk how to solve :(

OpenStudy (unklerhaukus):

factor the equation using prefect squares

OpenStudy (anonymous):

which equation??

OpenStudy (anonymous):

like the one of the circle?? (which i find??) or the general form?

OpenStudy (unklerhaukus):

\[x^2+ax+y^2+by+c=0\] \[\left(x+\frac a2\right)^2-a^2+y^2+by+c=0\]

OpenStudy (anonymous):

why whoops?? haha :P am i not seeing something ur seeing??

OpenStudy (unklerhaukus):

(i forgot to square ) (fixed now)

OpenStudy (anonymous):

ohhh okay i see :P so what do i do now?

OpenStudy (unklerhaukus):

factor the y terms using the same technique , then move the three constants outside the brackets to the right hand side

OpenStudy (anonymous):

(x+a/x)^2=(-a+y+by)^2 ?? did i do that right?? :/ idk...

OpenStudy (anonymous):

oops i forgot the +c... so like this?? (x+a/x)^2=(-a+y+by)^2+c ??? idk :(

OpenStudy (unklerhaukus):

\[\left(x+\frac a2\right)^2-a^2+y^2+by+c=0\] \[\left(x+\frac a2\right)^2-a^2+\left(y+\frac b2\right)^2-b^2+c=0\] \[\left(x+\frac a2\right)^2+\left(y+\frac b2\right)^2=a^2+b^2-c\]

OpenStudy (anonymous):

oh okay :P but how do i solve for a?

OpenStudy (unklerhaukus):

compair to the general form of a circle

OpenStudy (anonymous):

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