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Haven't done circle for long :( In the figure, a circle S: \(x^2+y^2+6x-2y=0\) cuts the axes at the origin O, A and C. A point B lies on the circle S so that the length of BC is \(2 \sqrt{5}\) units. BC is produced to cut the x-axis at D. Find the coordinates of B

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solve these equations for \( x_1, y_1 \), \[ x_1^2 + y_1^2 + 6x_1−2y_1=0,\\ x_2^2 + y_2^2 + 6x_2 - 2y_1 = 0, x_2=0, y_2 \neq 0 \\ (x_1-x_2)^2 + (y_1-y_2)^2 = 20\]

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Other answers:

May I know what (x2, y2) is??
x_2, y_2 are the coordinates of C
probably you would end up with two sets of point, choose the the point where y>0
sorry ... this equation .. change y1 to y2 \[ x_2^2 + y_2^2 + 6x_2 - 2y_1 = 0, x_2=0, y_2 \neq 0 \\ \]
Algebra matters.
C = (0, 2)
yes!! sorry i was careless ... did you find out answer?
\(x^2 + y^2 +6x - 2y =0\) ---(1) \(x^2+(y-2)^2 =20\) => \(x^2 + y^2 -4y+4 =20\) ---(2)
Do you mean I knew the answer?
I knew the answer, yes. I'm looking for a way to get the answer. I tried once (more than once) but I thought I was making it too complicated. I just wanted to find a simpler way so that my sister can do it easily.
co-ordinates of A can also be found similarly.
A = (-6, 0)
D = (4, 0)
then u know equation of line BD can't u use that to find co-ordinates of B ?
solving equation of line and circle simultaneously.
are u sure about D =(4,0)??
equation of line BD: \[\frac{y-2}{x} = \frac{2}{-4}\]\[x = 4-2y\] \[(4-2y)^2 + y^2 +6(4-2y) - 2y =0\] \[5y^2 - 30y + 40=0\]\[y=4, 2\]When y=4, x = 4-2(4) = -4 When y=2, x=4-2(2) =0 (rejected) So, B= (-4, 4) Yay! Much simpler :) Thanks!!! @hartnn Yes! Yes! D = (4,0)
oh, i somehow got 36 instead of 40, thats why i was stuck.....
Correction: o.O Did I make a mistake? :S
no, u are correct.
Okay, thanks again! I didn't thought of the line. I'm just too stupid :(
happens! i just did a silly mistake of 6*4 =20 :P
did anyone notice : |dw:1356789155290:dw|
where thats the center we know co-ordinates of C let center be O hence we know slope of OB slope given, distance give, we can find co-ordinate! B)
how do we know slope of OB?
as we know slope of OC
given slope of OB and O, we can find B, no need of any distance..
Not so relevant. |dw:1356789503661:dw|
ohh yes!! COA are collinear !
how does that make it any simpler ?
i dont know,,just a good fact to know! :D
''Not so relevant.'' :P
:P ok ..
u got 2 more easy methods ;)
Yes! Yes! Thanks! Thanks!!

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