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Find equation of a cricle

  1. Apr 29, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the equation of a cricle if it touches x axis in A(3,0) and it contains B(3 + √3, -1)

    3. The attempt at a solution
    Is there a piece of data missing here? Because i cant see how can i find center of circle.
     
  2. jcsd
  3. Apr 29, 2013 #2

    Ray Vickson

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    You need to show your work. For example, can you say ANYTHING about the center of the circle?
     
  4. Apr 29, 2013 #3
    Well i know that center is at x=3.

    I know that the slope between A and B is -√3 (3-3-√3)/(1)

    If i want to find a line that goes throug the points it should be like this:

    For A:
    y-0=-√3(x-3)
    y=-√3x + 3√3

    but for B i get different line

    y+1=-√3(x-3 -√3)
    y=-√3x + 3√3 +2

    Two different lines, when i should get the same.
     
  5. Apr 29, 2013 #4

    Mentallic

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    Nope, you have enough information. The circle touching the x axis means that the x-axis is a tangent to the circle. At the point a tangent line touches the circle, a line perpendicular to the tangent line (a normal line) would pass through the centre of the circle, would it not? This doesn't mean we know where the centre is exactly, but you know the centre must lie on the normal line. So can you use this added information in your equations?
     
  6. Apr 29, 2013 #5

    Mentallic

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    The only way the slope between A and B will be helpful to you is if you find the midpoint of A and B, take the perpendicular of the line AB going through the midpoint, then you'll know when this line intersects x=3, that is where your centre of the circle is. But there is an easier method to solve this problem.


    You found the gradient of the line that passes through A and [itex]C(3+\sqrt{3},-3)[/itex] which also lies on the circle (you shouldn't know this though). You accidentally took the slope to be the difference of the x values over the difference of the y values, but it's the other way around. The slope between A and B is

    [tex]\frac{-1-0}{3+\sqrt{3}-3}=\frac{-1}{\sqrt{3}}[/tex]
     
  7. Apr 29, 2013 #6
    Thank you everyone i solved the problem.
     
  8. Apr 29, 2013 #7

    Mentallic

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    Well in that case, I'll show you a quicker method which you may also find easier.

    The equation of the circle is

    [tex](x-a)^2+(y-b)^2=r^2[/tex]

    And we need to determine the constants a,b,r. Firstly, we know that the centre of the circle lies on the lies x=3 because of reasons I stated in my previous post, so a=3.

    So our circle equation is now

    [tex](x-3)^2+(y-b)^2=r^2[/tex]

    We're also given that the point (3,0) lies on the circle, so if we plug this into the equation we get

    [tex](3-3)^2+(0-b)^2=r^2[/tex]

    [tex]b^2=r^2[/tex]

    This equation makes sense right? Because if b is 2 units away from x-axis, then the radius is also 2 units.

    Finally, our last point that lies on the circle is [itex]B(3+\sqrt{3},-1)[/itex] and plugging this into the circle equation gives

    [tex](3+\sqrt{3}-3)^2+(-1-b)^2=r^2[/tex]

    [tex](\sqrt{3})^2+(b+1)^2=r^2[/tex]

    [tex]3+b^2+2b+1=r^2[/tex]

    Since we know that [itex]b^2=r^2[/itex] then

    [tex]2b+4=0[/tex]

    [tex]b=-2[/tex]

    And thus, plugging b=-2 into [itex]b^2=r^2[/itex] gives us [itex]r^2=4[/itex].

    Of course, when you're doing this yourself things will fall together a lot faster, and all you're really doing is plugging values that you're given into the equation of the circle to find the constants. It's brainless work really, and the only part of the problem that requires you to really think to solve it is to realize what "touching the x-axis" actual means and what info you can extract from that statement.
     
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