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Check the equation of the circle

  1. Mar 27, 2005 #1
    check plz the equation of the circle

    the Center is C(-1,2) and the x intercept is 3

    I found the distance between the center and the point to be r^2=10

    and the midpoint of the two given points to be (1,1)

    therefore my final equation for this circle is

    [tex] (x-1)^2 + (y-1)^2 =10 [/tex]

    is my answer correct? Any objections? :tongue2:

    Another question Equation of a circle with center (4,-3) that passes through the point (2,1)

    for this question I got [tex] (x-4)^2 + (y+3)^2 =20 [/tex]

    Is this one right?
     
    Last edited: Mar 27, 2005
  2. jcsd
  3. Mar 27, 2005 #2
    Unfortunately not quite. You don't need to find any midpoint at all: the center is given.

    Also I don't like your radius. It should just be the distance between the two given points (which I calculate to be [itex]\sqrt{20}[/itex]).
     
  4. Mar 27, 2005 #3
    Your answer to the second question is correct.
     
  5. Mar 27, 2005 #4
    oops ok I got my new answer to be

    [tex] (x+1)^2 + (y-2)^2 = 20 [/tex]

    is this correct now can you also check the second question in the first post? :redface:
     
  6. Mar 27, 2005 #5
    Yep, they're both right now. :smile:
     
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