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-   -   Linear equations concerning tangents (https://www.askmehelpdesk.com/showthread.php?t=64481)

  • Feb 19, 2007, 10:43 PM
    sim0nz12345
    Linear equations concerning tangents
    Hi
    Can you please answer this question and show me the solution, cheers.

    Show that the line 3x-4y+15=0 is a tangent to the circle x^2+y^2=9

    Thanks
  • Feb 19, 2007, 11:27 PM
    Capuchin
    You need to prove that your two curves cross at only one point, that defines a tangent.

    How would you do this, do you think?
  • Feb 19, 2007, 11:38 PM
    sim0nz12345
    I'm not sure. I don't really understand the concept of tangents and the unit circle.
    I just need the solution to it then it will get easier.
  • Feb 20, 2007, 12:36 AM
    Capuchin
    I just told you what a tangent is, it's a line that crosses at only one point (not 2 or 0 points).

    You don't have a unit circle, you have a circle of radius 3, so you don't need to worry about that.
  • Feb 20, 2007, 07:54 AM
    asterisk_man
    1: find the point(s) of intersection between the two equations given
    if step 1 finds more or less than 1 point of intersection then the line isn't tangent to the circle and you can quit here
    2: find the slope of the circle at the point of intersection and the slope of the line
    if the two slopes are equal then you've got a tangent line, otherwise you do not.

    If you need help completing those steps please ask here.
  • Feb 20, 2007, 01:55 PM
    sim0nz12345
    I not entirely sure how to find the intersection of the first step.
    I know simultaneous equations is invloved but I don't know what to do with the squared terms.
  • Feb 20, 2007, 02:07 PM
    asterisk_man
    try solving the linear equation for one variable, substitute into the circle equation, solve for the remaining variable, substitute that value into the linear equation solved for 1 var to find the other variable.
  • Feb 20, 2007, 10:28 PM
    sim0nz12345
    Ok thanks
    I eventually found the answer to be a perfect square which means that there can only be one point of intersection, in a tangent.
    x=-9/5
    y=12/5
    What do I then do for the next step
  • Feb 21, 2007, 05:18 AM
    galactus
    Plug your answers into the circle equation. Do you get 9? Plug them into the linear equation. Do you get 0?
  • Feb 21, 2007, 01:53 PM
    sim0nz12345
    Yes it does.
    Then what is the next step?
  • Feb 21, 2007, 02:09 PM
    galactus
    Next step? That's it.
  • Feb 21, 2007, 02:19 PM
    asterisk_man
    OK. I guess I agree. I was wrong. You don't need to compare the slopes. The only way that a line can have a single point of intersection with a circle is if it is tangent to the circle at that point. I agree with galactus. Your math is done. You just need to explain what I just said.

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