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Mathematics 16 Online
OpenStudy (ria23):

What is the value of b for the following circle in general form? x^2+y^2+ax+by+c=0

OpenStudy (ria23):

jimthompson5910 (jim_thompson5910):

The center of this particular circle is (1,-2) so h = 1 and k = -2

jimthompson5910 (jim_thompson5910):

the radius is 3, so r = 3

jimthompson5910 (jim_thompson5910):

plug these values into (x-h)^2 + (y-k)^2 = r^2 and try to get the equation into the form x^2+y^2+ax+by+c=0

OpenStudy (ria23):

\[(x-1)^{2}+(y+2)^{2}=\sqrt3\] I'm not really sure how to get it into that form. >~< \[x^{2}+2y^{2}+x+2y+3=0\]? ;-;

jimthompson5910 (jim_thompson5910):

notice how 'by' can only come from expanding out (y+2)^2 so you just need to expand that out using FOIL

OpenStudy (ria23):

y+2 y+2?

jimthompson5910 (jim_thompson5910):

(y+2)(y+2) = ??

OpenStudy (ria23):

Yea that

OpenStudy (ria23):

WAIT

OpenStudy (ria23):

Sorry, AFK \[y^{2}+4y+4\]?

jimthompson5910 (jim_thompson5910):

so by = 4y which means b = 4

OpenStudy (ria23):

:O And that's my answer... c: ?

jimthompson5910 (jim_thompson5910):

yes it is

OpenStudy (ria23):

Ahh. Thank yhu! ^_^

jimthompson5910 (jim_thompson5910):

you're welcome

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