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Callisto

  • 3 years ago

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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  1. Callisto
    • 3 years ago
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    |dw:1356781451262:dw|

  2. Callisto
    • 3 years ago
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    brb

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

  4. Callisto
    • 3 years ago
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    May I know what (x2, y2) is??

  5. experimentX
    • 3 years ago
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    x_2, y_2 are the coordinates of C

  6. experimentX
    • 3 years ago
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    probably you would end up with two sets of point, choose the the point where y>0

  7. experimentX
    • 3 years ago
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    sorry ... this equation .. change y1 to y2 \[ x_2^2 + y_2^2 + 6x_2 - 2y_1 = 0, x_2=0, y_2 \neq 0 \\ \]

  8. Callisto
    • 3 years ago
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    Algebra matters.

  9. Callisto
    • 3 years ago
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    C = (0, 2)

  10. experimentX
    • 3 years ago
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    yes!! sorry i was careless ... did you find out answer?

  11. Callisto
    • 3 years ago
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    \(x^2 + y^2 +6x - 2y =0\) ---(1) \(x^2+(y-2)^2 =20\) => \(x^2 + y^2 -4y+4 =20\) ---(2)

  12. Callisto
    • 3 years ago
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    he?

  13. Callisto
    • 3 years ago
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    Do you mean I knew the answer?

  14. Callisto
    • 3 years ago
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    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.

  15. hartnn
    • 3 years ago
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    co-ordinates of A can also be found similarly.

  16. Callisto
    • 3 years ago
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    A = (-6, 0)

  17. Callisto
    • 3 years ago
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    D = (4, 0)

  18. hartnn
    • 3 years ago
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    then u know equation of line BD can't u use that to find co-ordinates of B ?

  19. hartnn
    • 3 years ago
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    solving equation of line and circle simultaneously.

  20. hartnn
    • 3 years ago
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    are u sure about D =(4,0)??

  21. Callisto
    • 3 years ago
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    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)

  22. hartnn
    • 3 years ago
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    oh, i somehow got 36 instead of 40, thats why i was stuck.....

  23. Callisto
    • 3 years ago
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    Correction: o.O Did I make a mistake? :S

  24. hartnn
    • 3 years ago
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    no, u are correct.

  25. Callisto
    • 3 years ago
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    Okay, thanks again! I didn't thought of the line. I'm just too stupid :(

  26. hartnn
    • 3 years ago
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    happens! i just did a silly mistake of 6*4 =20 :P

  27. shubhamsrg
    • 3 years ago
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    did anyone notice : |dw:1356789155290:dw|

  28. shubhamsrg
    • 3 years ago
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    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)

  29. hartnn
    • 3 years ago
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    how do we know slope of OB?

  30. hartnn
    • 3 years ago
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    ohh..

  31. hartnn
    • 3 years ago
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    as we know slope of OC

  32. shubhamsrg
    • 3 years ago
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    yep..exactly..

  33. hartnn
    • 3 years ago
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    given slope of OB and O, we can find B, no need of any distance..

  34. Callisto
    • 3 years ago
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    Not so relevant. |dw:1356789503661:dw|

  35. shubhamsrg
    • 3 years ago
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    ohh yes!! COA are collinear !

  36. hartnn
    • 3 years ago
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    how does that make it any simpler ?

  37. shubhamsrg
    • 3 years ago
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    i dont know,,just a good fact to know! :D

  38. Callisto
    • 3 years ago
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    ''Not so relevant.'' :P

  39. hartnn
    • 3 years ago
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    :P ok ..

  40. hartnn
    • 3 years ago
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    u got 2 more easy methods ;)

  41. Callisto
    • 3 years ago
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    Yes! Yes! Thanks! Thanks!!

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