Results 1 to 4 of 4

Math Help - Circle Question

  1. #1
    Newbie
    Joined
    Jul 2007
    Posts
    12

    Circle Question

    Graph the circle (x+2)^2 + (y-1)^2= 25
    Last edited by charrie berri; August 3rd 2007 at 12:49 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by charrie berri View Post
    Graph the circle (x+2)^2 + (y-1)^2= 25

    Determine the equations of the tangents with slope of 3/4 =S
    Hello,

    1. It is a circle with the centre C(-2, 1) and the radius 5.

    2. The radius is perpendicular to the tangent in the tangent point. Thus the radius has the slope -4/3 and has to pass the centre:

    r: y=-\frac{4}{3}x-\frac{5}{3}

    Calculate the intersection of this line and the circle and you'll get the tangent points:

    (x+2)^2+\left(-\frac{4}{3}x-\frac{5}{3}-1\right)^2 = 25. After a few transformation you get:
    \frac{25}{9}x^2 + \frac{100}{9}x - \frac{125}{9}=0~\Longrightarrow~ x^2+4x-5=0. Solve for x. Plug in the values into the equation of the straight line r. The tangent points are:
    T_1(1, -3),~~T_2(-5, 5)

    3. You know the slope of the tangent and one point of the tangent. Use the point-slope-formula.
    t_1:y = \frac{3}{4}x-\frac{15}{4}

    t_2:y = \frac{3}{4}x+\frac{35}{4}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by charrie berri View Post
    Graph the circle (x+2)^2 + (y-1)^2= 25
    Hello,

    here is another approach:

    You allways find 2 parallel tangents to a circle. The equation of these tangents differ only in the constant summand. With your values the tangents have the form:

    t: y=\frac{3}{4}x + C

    Now calculate the intersection points of these tangents and the circle. Plug in the tangent term into the equation of the circle:

    (x+2)^2+(\frac{3}{4}x + C-1)^2=25 . Expand the brackets and after a few steps you'll get the equation:

    25 \cdot x^2  + 8 \cdot x \cdot (3 \cdot C + 5) + 16C  - 32C = 320. Solve for x:

    x=\frac{4\left(\sqrt{-16C^2+80C+525}-3C-5\right)}{25}~~\vee~~ ~~ x=-\frac{4\left(\sqrt{-16C^2+80C+525}-3C-5\right)}{25}

    You get only one intersetion point (that is the tangent point) if the radicand equals zero:

    \sqrt{-16C^2+80C+525}=0. Solve for C. You'll get:

    C=\frac{35}{4}~~\vee~~C=-\frac{15}{4} Plug in this value into the equation of the parallel lines and you have the result you know from my previous post.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by charrie berri View Post
    Graph the circle (x+2)^2 + (y-1)^2= 25
    Hello,

    in my previous post I mentioned that the centre of the circle is C(-2, 1) and the radius is 5. I have some difficulties to imagine what difficulties you could have to draw a circle if all data are present

    But nevertheles here it is:

    By the way: Don't delete your problem when you have received an answer. Other members of the forum could benefit from your problem and the way it is solved.
    Attached Thumbnails Attached Thumbnails Circle Question-zweitangenten_kreis.gif  
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Circle question.
    Posted in the Geometry Forum
    Replies: 1
    Last Post: February 9th 2011, 07:07 PM
  2. Circle question
    Posted in the Geometry Forum
    Replies: 1
    Last Post: December 8th 2009, 06:25 AM
  3. Circle question
    Posted in the Geometry Forum
    Replies: 3
    Last Post: February 4th 2009, 11:52 PM
  4. circle question
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: December 9th 2008, 01:24 PM
  5. Circle Question
    Posted in the Geometry Forum
    Replies: 2
    Last Post: December 4th 2008, 01:44 PM

Search Tags


/mathhelpforum @mathhelpforum