Results 1 to 6 of 6

Math Help - finding the equation of the tangent line to a circle

  1. #1
    Junior Member
    Joined
    Dec 2009
    Posts
    71

    finding the equation of the tangent line to a circle

    seeing that r^2(1+m^2)=b^2...use this method to find an equation of the tangent line to the circle x^2+y^2=9 at the point ( 1, 2square root of 2 )

    i'm totally lost...thanks in advance
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by alessandromangione View Post
    seeing that r^2(1+m^2)=b^2...use this method to find an equation of the tangent line to the circle x^2+y^2=9 at the point ( 1, 2square root of 2 )

    i'm totally lost...thanks in advance

    I'm not familiar with the equation r^2(1+m^2)=b^2, but I can tell you that the slope of the tangent line is perpendicular to the slope of the radius to the point (1 , 2\sqrt{2})

    slope of the radius is \frac{2\sqrt{2}}{1}

    tangent slope would be -\frac{1}{2\sqrt{2}}

    tangent line equation is ...

    y - 2\sqrt{2} = -\frac{1}{2\sqrt{2}}(x-1)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Dec 2009
    Posts
    71
    Quote Originally Posted by skeeter View Post
    I'm not familiar with the equation r^2(1+m^2)=b^2, but I can tell you that the slope of the tangent line is perpendicular to the slope of the radius to the point (1 , 2\sqrt{2})

    slope of the radius is \frac{2\sqrt{2}}{1}

    tangent slope would be -\frac{1}{2\sqrt{2}}

    tangent line equation is ...

    y - 2\sqrt{2} = -\frac{1}{2\sqrt{2}}(x-1)
    well the result should be square root of 2 +4y-9square root of 2=0....maybe someone else can help me out?
    anyway thank u !
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Apr 2010
    Posts
    25

    I need help! ):

    I am completely stuck on this problem. please help me work it out. ):

    Find a formula that gives the slope of a line tangent to the circle (x^2)+(y^2)=(r^2) at any point (in the first quadrant) (x,sqrt((r^2)-y^2)).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Jun 2009
    Posts
    806
    Thanks
    4
    Quote Originally Posted by alessandromangione View Post
    seeing that r^2(1+m^2)=b^2...use this method to find an equation of the tangent line to the circle x^2+y^2=9 at the point ( 1, 2square root of 2 )

    i'm totally lost...thanks in advance
    In the given problem r = 3.
    Differentiating the equation of the circle, you get m. Thus
    2x + 2y*y'= 0 Put the values of x and y and find y' = m.
    Using the formula
    r^2(1+m^2) = b^2, find b.
    Then the equation of the tangent is
    y = mx + b.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by alessandromangione View Post
    seeing that r^2(1+m^2)=b^2...use this method to find an equation of the tangent line to the circle x^2+y^2=9 at the point ( 1, 2square root of 2 )

    i'm totally lost...thanks in advance
    1. If your equation is satisfied by r, m and b the line in question is a tangent (in Germany this equation is called "condition for the existance of a tangent to a circle")

    2. The tangent point is T\left(\frac{-bm}{1+m^2} , \frac{b}{1+m^2}\right)

    3. Consequently you'll get a system of equations:

    \left|\begin{array}{rcl}\dfrac{-bm}{1+m^2}&=&1 \\ \\ \dfrac{b}{1+m^2}&=&2\sqrt{2} \end{array}\right.

    4. Solve this system for b and m and plug in these values into the equation of a straight line.
    Spoiler:
    I've got y = -\frac14 \sqrt{2} \cdot x + \frac94 \sqrt{2}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: January 12th 2011, 02:38 PM
  2. Replies: 2
    Last Post: October 12th 2010, 02:24 PM
  3. Replies: 1
    Last Post: June 8th 2010, 04:44 PM
  4. finding equation of tangent line
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 12th 2010, 06:07 AM
  5. Finding equation of a tangent line.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 5th 2008, 07:20 PM

Search Tags


/mathhelpforum @mathhelpforum