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Derive the general formula of the equation of a circle for the points

  1. Jul 24, 2013 #1
    1. The problem statement, all variables and given/known data

    1. (a) Find the equation of the circle with the straight line joining A(1;-1) and C(3; 4)
    as diameter.
    (b) Hence or otherwise, derive the general formula
    (x - x1)(x - x2) + (y - y1)(y - y2)=0
    of the equation of a circle for the points A(x1; y1) and C(x2; y2):

    2. Relevant equations



    3. The attempt at a solution

    I did part a) but part b) is baffling me could someone give me a hint on what to do?
     
    Last edited: Jul 24, 2013
  2. jcsd
  3. Jul 24, 2013 #2

    Mark44

    Staff: Mentor

    Something missing here? I would think that a "general formula" would be an equation, which the above is not. Is this the complete problem statement?
     
  4. Jul 24, 2013 #3
    sorry it was supposed to be equal to 0
     
  5. Jul 24, 2013 #4

    LCKurtz

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    If you did part a, do part b the same way using the variables instead of the numbers. Then see if you can get it in the required form.

    [Edit] Alternatively use this hint: Those points and the point (x,y) can be used to make an angle inscribed in a semicircle, so...
     
  6. Jul 24, 2013 #5
    Yeah I tried and took the gradient of the lines and got it. Thanks.
     
  7. Jul 25, 2013 #6

    HallsofIvy

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    Impossible-this is NOT the equation of a circle!

     
  8. Jul 25, 2013 #7
    It looks like the solution to me

    [tex] (x-x_1)(x-x_2)+(y-y_1)(y-y_2)=x^2-(x_1+x_2)x+x_1x_2+y^2-(y_1+y_2)y+y_1y_2 \\
    =(x-\frac{x_1+x_2}{2})^2-(\frac{x_1+x_2}{2})^2+x_1x_2+(y-\frac{y_1+y_2}{2})^2-(\frac{y_1+y_2}{2})^2+y_1y_2=0 \\

    (x-\frac{x_1+x_2}{2})^2+(y-\frac{y_1+y_2}{2})^2= \frac{x_1^2+2x_1x_2+x_2^2}{4}-x_1x_2+\frac{y_1^2+2y_1y_2+y_2^2}{4}-y_1y_2

    \\

    (x-\frac{x_1+x_2}{2})^2+(y-\frac{y_1+y_2}{2})^2= \frac{x_1^2-2x_1x_2+x_2^2}{4}+\frac{y_1^2-2y_1y_2+y_2^2}{4}
    \\
    (x-\frac{x_1+x_2}{2})^2+(y-\frac{y_1+y_2}{2})^2= \frac{(x_1-x_2)^2+(y_1-y_2)^2}{4} [/tex]

    This looks like the equation of a circle with the right center and radius.
     
  9. Jul 25, 2013 #8

    ehild

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    The centre of the circle is the midpoint of the diameter, at O((x1+x2)/2;(y1+y2)/2). The radius is half of the diameter: R=0.5sqrt((x2-x1)2+(y2-y1)2). Writing up the equation of a circle with these parameters, arranging, factorising and simplifying, you get the desired formula.

    ehild
     
  10. Jul 25, 2013 #9

    LCKurtz

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    Lionely seemed to get my hint. The equation is trivially correct if you note that if ##P=(x,y)## is a point on the circle with ##A=(x_1,y_1),B=(x_2,y_2)## then angle ##APB##, being inscribed in a semicircle, is a right angle. The equation is simply ##\vec{AP}\cdot \vec{BP}=0##.
     
  11. Jul 25, 2013 #10

    ehild

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    That is the nicest solution! :smile:

    ehild
     
  12. Jul 25, 2013 #11
    I didn't use vectors I took the gradient but your solution was like 2 lines mine was like 5 or more . Using vectors like that is ... awesome.
     
  13. Jul 25, 2013 #12

    symbolipoint

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    I'm confused what you want in part (b). The general equation for circle in two dimensions is
    [itex](x-a)^2+(y-b)^2=r^2[/itex]
    and comes from using the distance formula for an arbitrary center of the circle.

    If you would take your two given points, use middle-point formula, this would be your circle's center. Half the length of the segment joining your two given points has length of your radius.
    Use Distance formula with the general point, (x,y), the express distance from center to this general (x,y), set it equal to r, and then derive your equation.
     
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