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Coordinate geometry - circles.

  1. Aug 11, 2004 #1
    I'm just introducing myself to coordinate geometry in the xy plane of cirlces.
    Here's a question i'm having trouble with:

    Q11: The line with equation y=mx is a tangent to the circle with equation x^2 + y^2 - 6x - 6y + 17 = 0. Find the possible values of m.

    At first i thought i'd try substituting y=mx into the curve equation, but i was still left with 2 unknowns. I don't really know what to do here. Could anyone offer some advice? Thanks.
    The answer is (9 (+ or -) root 17) / 8
     
  2. jcsd
  3. Aug 11, 2004 #2
    After substituting you should have

    [tex]x^2-m^2x^2-6x-6mx+17=0[/tex]

    Rearrange this and you get

    [tex](1+m^2)x^2 - (6m+6)x + 17 = 0[/tex]

    Let a = 1 + m2, b = -(6m + 6) and c = 17. Then use the quadratic equation (and remember that you want the expression under the square root to be greater than or equal to 0).
     
    Last edited: Aug 11, 2004
  4. Aug 11, 2004 #3
    Your latex looks ok. Do you mean rearrange to form:
    x^2 ( 1+ m^2) - x(6m+1) + 17 = 0 ?

    Edit: Ah yes you typed that below. I'll try that now.
     
  5. Aug 11, 2004 #4
    I re-edited my post. Please have a look.
     
  6. Aug 11, 2004 #5
    I get it. So you say that b^2 - 4ac = 0. Then form yet another quadratic equation, tidy up and simplify. I haven't had much experience of using brackets as coefficients in the quadratic equation but that's something i'm going to remember for the future.

    Thanks for the help!
     
  7. Aug 11, 2004 #6
    That should be b2 - 4ac ≥ 0.
     
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