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Check Eqn of Circle with diameter end points(1,2) and (3,-6)

  1. Mar 26, 2005 #1
    Check plz Eqn of Circle with diameter end points(1,2) and (3,-6)

    Find the equation of a circle with a diameter that has end points P(1,2) and Q(3,-6)

    I know the equation for a circle in general form is [tex]x^2+y^2=r^2 [/tex]

    First I found the distance between the two points given because this also equals the diameter.

    I used point P as x1,y1 and Q as x2,y2 and subbed these points into the distance equation.

    I got Distance=Diameter=sqrt(68) I took half of this to get the radius. 5.83 and then I put this into the equation for a circle and squared it to get 33.99

    My final equation of a circle for this question is [tex] x^2+y^2=34 [/tex]

    Can someone please check my answer and tell me if I went wrong anywhere? :smile:
     
  2. jcsd
  3. Mar 26, 2005 #2
    if the equation is [tex] x^2+y^2=34 [/tex], then the centre is (0,0). but according to your diameter, the centre should not be (0,0). btw, the radius is 4.1231...

    i think the equation is (x-2)^2+(y+2)^2=17
     
    Last edited: Mar 26, 2005
  4. Mar 26, 2005 #3
    um using the points I found the midpoint to be (1,-4) does that mean this equation should be written as

    [tex](x-1)^2 + (y+4)^2 = 34 [/tex]
     
    Last edited: Mar 26, 2005
  5. Mar 26, 2005 #4
    no....[tex] (x-1)^2+ (y+4) ^2= 34 [/tex]
     
  6. Mar 26, 2005 #5
    lol....i graphed both equations (mine and yours) in my graphic calculator...both are wrong...
     
  7. Mar 26, 2005 #6
    I was just gonna write the answer that you wrote I remember I standard form, are you sure it is wrong? Maybe only the 34 part is wrong I had to do some rounding.

    (x-2)^2+(y+2)^2=17 is this one right?
     
    Last edited: Mar 26, 2005
  8. Mar 26, 2005 #7
    acutally, i really think your radius is wrong, sqrt 68 then divide by 2 is 4.1231. i also think that the transformation part is right , but it turns out to be wrong in my calculator
     
  9. Mar 26, 2005 #8
    "(x-2)^2+(y+2)^2=17 is this one right?"

    i think so. but my graphic calc says i am wrong
     
  10. Mar 26, 2005 #9
    yes my radius was wrong now I think the equation is

    [tex] (x-1)^2 + (y+4)^2 = 17 [/tex] what do you think? this seems right

    the coordinates of the center will be the midpoint of the two points which is (1,-4)
     
  11. Mar 26, 2005 #10
    nope, it's even more wrong then mine
     
  12. Mar 26, 2005 #11
    why you think the radius is (1,-4)

    i think its (2,-2)
     
  13. Mar 26, 2005 #12
    I dont think that is the radius I think that is the center point (h,k)

    The question says find the equation of a circle the diamter has end points P(1,2) and Q(3,-6) This is the diameter the middle of the diamter will be the center thats why im taking the midpoint as (1,-4)

    [(x2-x1)/2, (y1-y2)/2]
     
  14. Mar 26, 2005 #13
    think about this, the diameter given is sloped, thus the centre of the diameter wouldnot have the same x value as one of its end points...
     
  15. Mar 27, 2005 #14
    i think my graphic calc is wrong (since i rememebr my teacher told me that he might accidently changed my program). just copy my equation.
     
  16. Mar 27, 2005 #15
    ok tell me where you got ur point (2,-2)?
     
  17. Mar 27, 2005 #16
    The midpoint formula is [tex]\left( \frac{x_1+x_2}{2}, \ \frac{y_1+y_2}{2}\right)[/tex], not [tex]\left( \frac{x_2-x_1}{2}, \ \frac{y_2-y_1}{2} \right)[/tex].
     
  18. Mar 27, 2005 #17
    so is my equation wrong data? because i can't find out where i got wrong.
     
    Last edited: Mar 27, 2005
  19. Mar 27, 2005 #18
    I believe [itex](x-2)^2 + (y+2)^2 = 17[/itex] is correct, yes~

    By the way, there's no need to convert anything to decimal form for this question, or for most others. Your radius is [itex] r[/itex] in

    [tex] r^2 = \left(\frac{\sqrt{68}}{2}\right)^2 = \frac{68}{4} = 17.[/tex]
     
  20. Mar 27, 2005 #19

    dextercioby

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    Because the coordinates of the center of the circle must be in certain relation with the ones of the ends of the diameter.Use the symmetry of the circle to get the correct equation as

    [tex] (x-2)^{2}+(y+2)^{2}=17 [/tex]

    U can check that 2 points are on that circle &,moreover,the radius is [itex] \sqrt{17} [/itex]...

    Daniel.
     
  21. Mar 27, 2005 #20
    hey thanks so much, my midpoint formula was wrong in my online text no wonder I wasnt getting the point (2,-2) but now everything is fine. :smile:
     
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