Results 1 to 8 of 8

Math Help - Tangent Line of a Circle

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    20

    Tangent Line of a Circle

    Hey guys, I've tried this one a few times and just can't seem to get it. Help would be appreciated:

    A circle has its center at C(4,8). A point Q(12,6) is on the circumference of the circle. A tangent line to the circle is drawn at point Q. Determine the equation of the tangent line.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Vesicant View Post
    Hey guys, I've tried this one a few times and just can't seem to get it. Help would be appreciated:

    A circle has its center at C(4,8). A point Q(12,6) is on the circumference of the circle. A tangent line to the circle is drawn at point Q. Determine the equation of the tangent line.
    Since you originally posted this in the Pre-algebra and Algebra subforum (and I moved it to here) I assume that you're to do this without using calculus ..... Correct?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2009
    Posts
    20
    Quote Originally Posted by mr fantastic View Post
    Since you originally posted this in the Pre-algebra and Algebra subforum (and I moved it to here) I assume that you're to do this without using calculus ..... Correct?
    Correct
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Vesicant View Post
    Correct
    Draw a diagram. Note the fact that the angle between the tangent and the radius is 90 degrees. Use this fact, geometry and some simple trigonometry to get the angle \theta that the tangent makes to the x-axis. Then m = \tan \theta.

    Knowing the gradient of a line and a point on it, you should be able to get the equation of the line.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,201
    Thanks
    1789
    You might also use the fact that if the lines corresponding to y= m_1x+ b_1 and y= m_2 x+ b_2 are perpendicular, then m_1m_2= -1. You know that a radius passing through (4, 8) and (12, 6). What is the slope of that line? You know that the tangent at (12, 6) is perpendicular to that radius so you can find the slope of the tangent. Now you know the slope and one point of the line so you can find the equation of the line.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Sep 2009
    Posts
    20
    Quote Originally Posted by HallsofIvy View Post
    You might also use the fact that if the lines corresponding to y= m_1x+ b_1 and y= m_2 x+ b_2 are perpendicular, then m_1m_2= -1. You know that a radius passing through (4, 8) and (12, 6). What is the slope of that line? You know that the tangent at (12, 6) is perpendicular to that radius so you can find the slope of the tangent. Now you know the slope and one point of the line so you can find the equation of the line.
    So if the slope given by (4,8) and (12,6) is M=8-6/4-12=2/-8=-1/4 then the slope of the tangent is 4/1 which would make the equation of the tangent Y=4/1x+3?

    Here's a quick graph I did of it, Something doesn't feel right though :/

    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,201
    Thanks
    1789
    How did you get "+3"? Yes, the slope is 4 so the equation is of the form y= 4x+ b.
    In order that it pass through (12, 6) you must have 6= 4(12)+ b so b= 6- 48= -42.
    Try y= 4x- 42.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Sep 2009
    Posts
    20
    Quote Originally Posted by HallsofIvy View Post
    How did you get "+3"? Yes, the slope is 4 so the equation is of the form y= 4x+ b.
    In order that it pass through (12, 6) you must have 6= 4(12)+ b so b= 6- 48= -42.
    Try y= 4x- 42.
    Yeah I know what I did wrong lol, Thought about that while sitting in class today for some reason I put the X intercept into the equation haha.

    So yes [Math]Y=4x-42[/tex] is what I got after reworking it, Thanks for all the help.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: August 19th 2011, 07:59 AM
  2. Circle, tangent line, and a point not on the circle
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: March 31st 2011, 02:40 PM
  3. Replies: 1
    Last Post: June 8th 2010, 05:44 PM
  4. finding the equation of the tangent line to a circle
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: April 13th 2010, 11:20 AM
  5. Circle tangent to the line.
    Posted in the Geometry Forum
    Replies: 11
    Last Post: January 15th 2010, 10:09 AM

Search Tags


/mathhelpforum @mathhelpforum