1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Circle problem

  1. Jan 5, 2008 #1
    Circle C has equation [itex] x^2+y^2+6x-16=0 [/itex]

    i) find centre and radius (turned out to be centre (-3,0) and radius 5)
    ii)verify that point A, (0,4) lies on C
    iii)Find the coords of point D, given that AD is a diameter of C

    I can do parts i) and ii) and for part iii) would I just use a simultaneous equation with the equation of the line from centre to point A (which worked out as 4/3x+4) and the equation of the circle? Because it seems a bit long winded
     
  2. jcsd
  3. Jan 5, 2008 #2
    Point D is at the other end of the circle, with respect to A. How would you express its coordinates?
     
    Last edited: Jan 5, 2008
  4. Jan 5, 2008 #3
    I can work it out easily if i sketch the circl;e and it comes out as (-6,-4), but there must be some kind of method to obtain the answer without the sketch.
     
  5. Jan 5, 2008 #4
    Let me put it this way. Imagine a vector from the center to A. What is the vector from the center to D?
     
  6. Jan 5, 2008 #5
    You can subtract two points to obtain a vector:

    Make a vector [tex]\vec{OA}[/tex] by doing A-O, where O=(0,0), then actually [tex]\vec{OA}[/tex] becomes (0,4).

    Then let S be the centre of the circle, so S = (-3,0)
    And make a vector pointing from A to S [tex]\vec{AS}[/tex] = S - A = (-3,-4)

    Then just use add vectors like this:
    [tex]\vec{OA}[/tex] + [tex]\vec{AD}[/tex] = [tex]\vec{OD}[/tex]
    but because [tex]\vec{AD}[/tex] is diameter it means that:
    [tex]\vec{AD}[/tex] = 2*[tex]\vec{AS}[/tex]

    So it becomes:
    [tex]\vec{OA}[/tex] + 2*[tex]\vec{AS}[/tex] = [tex]\vec{OD}[/tex]
    From there you have [tex]\vec{OD}[/tex] = (-6,-4)
    And because [tex]\vec{OD}[/tex] = D - O it is the point D, and its coordinates are (-6,-4)
     
  7. Jan 5, 2008 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You could use congruent triangles. If A= ([itex]x_1[/itex],[itex]y_1[/itex]) is on the circle with center O= ([itex]x_0[/itex],[itex]y_0[/itex]), and AB is a diameter, with B= ([itex]x_2[/itex],[itex]y_0[/itex], then The triangle with vertices A, O, and ([itex]x_0[/itex],[itex]y_1[/itex]) is congruent to the triangle with vertices B, O, and ([itex]x_0[/itex], [itex]y_1[/itex]).
     
  8. Jan 7, 2008 #7

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    From the centre and radius you have worked out, express the equation of the circle in co-ordinates (x’, y) (you are lucky you have to change only one co-ordinate, the centre lies on the x axis) that have the centre as origin. From your values you know it has got to be (x + 3)^2 + y^2 = 25. Check that this is the same as your original equation. Call it
    x’^2 + y^2 = 25

    where x’ = (x + 3).

    What is the point diametrically opposite to a point A coordinates (x’, y) on a circle centred on origin? Easy -> (-x’, -y)

    Now you have that in the new co-ordinates, what is it expressed in the old ones? See last equation.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Circle problem
  1. Circle Problem ! (Replies: 11)

  2. 6 circles problem (Replies: 2)

  3. The circle problem (Replies: 22)

Loading...