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samnatha

please help i've some idea of how to do it but i'm not sure ? Find the equation of the circle which passes through the points P(-3,-2) and Q(0, -1) and has the line 2x - y + 4 =0 as a tangent at the point P(-3,-2)

  • one year ago
  • one year ago

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  1. phi
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    what's your idea?

    • one year ago
  2. samnatha
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    am well you use the x^2 + y^2 + 2gx + 2fy _ c =0 formula and sub the points into to get the two equations and then to get the third one you sub in -g for x and -f for y ?

    • one year ago
  3. phi
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    |dw:1355075215700:dw|

    • one year ago
  4. samnatha
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    i have no idea what your talking about sorry you kind of lost me

    • one year ago
  5. phi
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    I changed my mind. |dw:1355075548150:dw|

    • one year ago
  6. samnatha
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    yes i get that but how do i get the equation ?

    • one year ago
  7. phi
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    the line normal to y= 2x+4 has slope -1/2 and goes thru pt (-3,-2) the segment that goes through pts (-3,-2) and (0,-1) has slope (-1+2)/(0+3)= 1/3 the normal will have slope -3 (-1/m) it goes thru the midpt (-3/2, -3/2), so its equation is y= -3x-6

    • one year ago
  8. samnatha
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    could you go through that slower because we have been doing a different way than that ?

    • one year ago
  9. phi
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    maybe we should do it the way you are learning to do it. how far did you get?

    • one year ago
  10. samnatha
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    well i subed in all my points and got the three equations which are -6g - 4f + c =0 -2f + c =0 -2g + f = 0

    • one year ago
  11. phi
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    3 equations and 3 unknowns is good. I would use "elimination" to solve them

    • one year ago
  12. samnatha
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    i tried that and i got g = 2 but the answer in the back of the back of the book says its not that ?

    • one year ago
  13. phi
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    I think g should be 1 are you accounting for 2gx in front of x?

    • one year ago
  14. samnatha
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    how did you get that ?

    • one year ago
  15. samnatha
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    the answer in the back of the book says the equation is x^2 + y^2 + 2x + + 6y + 5 = 0

    • one year ago
  16. phi
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    yes, the book is correct.

    • one year ago
  17. samnatha
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    i din't get the radius or anything ?

    • one year ago
  18. phi
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    my way worked, but it is not the way to learn how to do your approach. I got (-1,-3) as the center and sqrt(5) for the radius

    • one year ago
  19. phi
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    in your approach you find g, f, and c and get the equation you posted x^2 + y^2 + 2x + 6y + 5 = 0 we could (if we had to) figure out the center of the circle and the radius by completing the square for x^2 and y^2. But first, how did you set up your equations?

    • one year ago
  20. samnatha
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    ok the first one for the point (-3. -2) i did this 9 + 4 + 2g(-3) + 2f(-2) + c = 0

    • one year ago
  21. samnatha
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    the second one (0, -1) i did 1 + 2g(0) + 2f(-1) + c =0

    • one year ago
  22. samnatha
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    and for the third i subed in (-g, -f) and i got this 2(-g) -y(-f) + 4 = 0

    • one year ago
  23. phi
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    for the 3rd equation I would would replace x with -g and y with -f to get \[g^2 + f^2 -2g^2-2f^2+c=0 \]

    • one year ago
  24. phi
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    that simplifies to \[ g^2 + f^2 -c=0 \]

    • one year ago
  25. samnatha
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    we never sub it into the formula usually

    • one year ago
  26. phi
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    where do you sub in (-g, -f)? what do you sub this into?

    • one year ago
  27. samnatha
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    the line 2x - y + 4 =0 to get the third equation

    • one year ago
  28. phi
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    that only makes sense if you use the line perpendicular to the tangent line (-g,-f) is the center of the circle. |dw:1355078200487:dw|

    • one year ago
  29. phi
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    the line you want has slope -1/2 and goes thru point (-3,-2) y+2 = (-1/2)(x+3)

    • one year ago
  30. samnatha
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    oooh right so how would i do it then ?

    • one year ago
  31. phi
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    they give you 2x - y + 4 =0 in slope-intercept form that is y= 2x+4 (just add y to both sides) its slope is 2. the slope of the line we want is -1/m or in this case -1/2

    • one year ago
  32. samnatha
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    how did you get the other slope ?

    • one year ago
  33. phi
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    the slope of lines that are perpendicular are "negative reciprocals" if one has slope m that other has slope -1/m

    • one year ago
  34. samnatha
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    ok i see it now you flip the slope upside down and change the signs

    • one year ago
  35. samnatha
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    so what would i do next ?

    • one year ago
  36. phi
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    now sub in (-g,-f) into the equation x+2y+7=0 now you have 3 eqs solve for g,f and c

    • one year ago
  37. samnatha
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    where did you get x + 2y + 7 =0 from ?

    • one year ago
  38. phi
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    After we figured out its slope was -1/2 we using the point slope formula or just write down y= mx+b and replace m with -1/2 , and (-3,-2) for (x,y) (the line goes thru this point) solve for b. then write the equation in standard form. -2 = (-1/2) * (-3) +b b= -2 - 3/2 = -7/2 y= (-1/2) x -7/2 multiply both sides by 2 2y = -x -7 add +x+7 to both sides x+2y+7 =0

    • one year ago
  39. samnatha
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    so do i do that with the other equations aswell ?

    • one year ago
  40. phi
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    your eqs are 9 + 4 + 2g(-3) + 2f(-2) + c = 0 <-- -4f -6g +c = -13 1 + 2g(0) + 2f(-1) + c =0 <-- -2f +c = -1 g+2f-7=0 <-------------------- 2f +g = 7

    • one year ago
  41. samnatha
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    ah ok i get it now :) thanks

    • one year ago
  42. phi
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    I got g=1, f=3, and c=5

    • one year ago
  43. phi
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    I noticed your original equations were all set = to 0 (and they should not have been) so that is one other thing to watch out for.

    • one year ago
  44. samnatha
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    ok thanks

    • one year ago
  45. phi
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    can you solve the 3 eqs?

    • one year ago
  46. samnatha
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    i'm just doing it now

    • one year ago
  47. samnatha
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    i got 19/5 for g but i don't think thats right ?

    • one year ago
  48. samnatha
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    wait it don't matter i see my mistake g is 1

    • one year ago
  49. phi
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    It is a pain doing 3 eq and 3 unknowns

    • one year ago
  50. samnatha
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    the i got 6 for f ? is that right ?

    • one year ago
  51. samnatha
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    oh no wait i got it haha sorry i make really stupid mistakes

    • one year ago
  52. samnatha
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    i got the answer thank you so much for the help :)

    • one year ago
  53. phi
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    no. Here is how I do it. arrange alphabetically (and it is good the coeff of c is 1) c f g constant 1 -4 -6 -13 <-- -4f -6g +c = -13 1 -2 0 -1 <---- -2f +c = -1 0 2 1 7 <------ 2f +g = 7 subtract replace row 2 with row 2 - row 1 1 -4 -6 -13 0 2 6 12 0 2 1 7 now replace 3rd row with 3rd row - 2nd row 1 -4 -6 -13 0 2 6 12 0 0 -5 -5 now back solve -5g= -5 so g= 1 2f +6(1)= 12 2f =6 f=3 c -4(3) -6(1)= -13 c -18 = -13 c= 5

    • one year ago
  54. samnatha
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    yupe i got all that thank you for your help would not have got it without you :)

    • one year ago
  55. phi
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    yw

    • one year ago
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