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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

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    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

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    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.
     
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