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Finding the equation of a circle from given point on the graph

  1. Oct 19, 2005 #1

    liz

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    Hi, im sure this question is really easy, but i don't have a clue how to do this question....

    Three points are P (-2, 7), Q (2,3), and R (4, 5). Find the equation of the circle which passes through points P, G, and R.

    Thanks lots to anyone who helps. liz x
     
  2. jcsd
  3. Oct 19, 2005 #2

    Galileo

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    Well, if you could find the center of the circle the rest is easy.
    Suppose you draw a circle and pick two arbitrary (different) points on that circle. What construction with only these points would allow you to draw a line that passes through the center of the circle?
    Once you know this, you can apply it to 2 pairs of points from P,Q,R and construct the center.
     
  4. Oct 19, 2005 #3

    TD

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    Or, an alternative method, you can consider the triangle formed by the 3 points. Then try to find the circumcircle.
     
  5. Oct 19, 2005 #4

    HallsofIvy

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    Those are very good geometric ways to specify the circle. Since the original question asked for the equation, you could also do this: the equation of a general circle with center (a,b) and radius R is
    (x- a)2+ (y- b)2= R2.

    Since the circle passes through (-2, 7), x=-2, y= 7 must satisfy that:
    (-2-a)2+ (7-b)2= R2.
    Do the same with the other two points and you have 3 equations to solve for the 3 unknowns a, b, and R.
     
  6. Oct 20, 2005 #5
    It always helps plot given points because the question might present a special case.

    It also helps a lot use graph paper.

    After you plot these points you might notice that PQ and QR go through the grid intersictions in a special way.
    Can you prove what you see? There is a certain rule you should apply.

    Since a triangle PQR is special, there is an interesting property of the center of its circumcircle. It makes it a snap to find its coordinates and radius, and equation of the circle.

    If you get stumped I'll be happy to help you along.
     
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