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Cartesian circles

  1. Oct 18, 2008 #1

    Mentallic

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    I have 3 circles, all cleverly plotted so as to touch the unit circle just once:

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

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

    [tex](x-\frac{\sqrt{7}}{2})^2+(y-\frac{1}{2})^2=3-2\sqrt{2}[/tex] (3)

    What I need is a circle (in general form like those above) that touches all 3 circles just once. This is possible just by looking at the graph of the 3 circles. I know that this 4th circle will have its centre somewhere along the line that is perpendicular and passing through the midpoint to the line passing through the centre of the circles (2) and (3). This equation comes to be:

    [tex]3x+(2-\sqrt{7})y=0[/tex]

    From here I'm stuck. Usually I would find the distance between the centres of 2 circles to find the radius of 1 circle if I know the other, but in this case the centre is variates depending on how large the radius is.
     
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  3. Oct 18, 2008 #2

    tiny-tim

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    Hi Mentallic! :smile:

    Why not just choose a general point on that line, and calculate the distances to the centres of the circles? :wink:
     
  4. Oct 20, 2008 #3

    Mentallic

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    Because I need to find the perfect distance (which is going to end up being some complicated surd) so that the circle will touch all 3 circles just once and perfectly. Maybe a diagram would be more appropriate in this case:

    [​IMG][​IMG]

    All the horizontal and vertical lines are just the exact intersections of the circles.

    Let [tex]x^2+y^2=1[tex] be circle #1

    [tex](x+1)^2+(y-1)^2=3-2\sqrt{2}[tex] be circle #2 (left-most circle)

    [tex](x-\frac{\sqrt{7}}{2})^2+(y-\frac{1}{2})^2=3-2\sqrt{2}[tex] be circle #3 (right-most circle)

    The intersection between circle #1 and #2 is:
    [tex](\frac{1}{2}[1+\sqrt{2}][\sqrt{3-2\sqrt{2}}-1],\frac{1}{\sqrt{2}}\sqrt{-\sqrt{2}+3\sqrt{3-2\sqrt{2}}+2\sqrt{6-4\sqrt{2}}})[tex]

    The intersection between circle #1 and #3 is:
    [tex](\frac{1}{4}[1+\sqrt{2}][\sqrt{7}-\sqrt{21-14\sqrt{2}}],\frac{1}{2\sqrt{2}}\sqrt{-6-7\sqrt{2}+7[3+2\sqrt{2}]\sqrt{3-2\sqrt{2}}})[tex]

    With the intersection of circles #1 and #3, the x-value of these seemed to intersect the perpendicular bisector [tex]3x+(2-\sqrt{7})y=0[tex] at roughly the spot where I expected the centre of the circle to be. Since I was stuck with my attempts to try and find a solution, I used this x-value as the centre of the circle and by finding the distances between the intersection of the perpendicular bisector and the top-edge of circle #1, with this suspected centre of the required circle, I ended up with the equation:

    [tex](x-\frac{1}{4}[1+\sqrt{2}][\sqrt{7}-\sqrt{21-14\sqrt{2}}])^2+(y-\frac{3}{4[\sqrt{7}-2]}[1+\sqrt{2}][\sqrt{7}-\sqrt{21-14\sqrt{2}}])^2=[\frac{1}{4}[1+\sqrt{2}][\sqrt{7}-\sqrt{21-14\sqrt{2}}]-\frac{\sqrt{3-\sqrt{7}}}{2\sqrt{2}}]^2+[\frac{3}{4[\sqrt{7}-2]}[1+\sqrt{2}][\sqrt{7}-\sqrt{21-14\sqrt{2}}]-\frac{\sqrt{5+\sqrt{7}}}{2\sqrt{2}}]^2[tex]

    But even though this circle is perfectly touching circle #1 just once also where the perpendicular bisector cuts circle #1, this circle doesn't touch circles #2 and #3 perfectly. They are ever-so slightly off. I need another approach to find the centre of the circle.

    EDIT: The latex images weren't appearing because they might be a little too long, so I removed the images and left them in text.
     
    Last edited: Oct 20, 2008
  5. Oct 23, 2008 #4

    Mentallic

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    I just began learning about taking the square root of complex numbers, so using the same idea I was able to simplify the coordinates of each intersection. Maybe it will help as I am still stumped on how to find the answer.

    Again:

    All the horizontal and vertical lines are just the exact intersections of the circles.

    Let [tex]x^2+y^2=1[/tex] be circle #1

    [tex](x+1)^2+(y-1)^2=3-2\sqrt{2}[/tex] be circle #2 (left-most circle)

    [tex](x-\frac{\sqrt{7}}{2})^2+(y-\frac{1}{2})^2=3-2\sqrt{2}[/tex] be circle #3 (right-most circle)

    The intersection between circle #1 and #2 is:
    [tex](\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}})[/tex]

    The intersection between circle #1 and #3 is:
    [tex](\frac{\sqrt{14}}{4},\frac{\sqrt{2}}{4})[/tex]

    I don't know how to go about finding the answer though, any thoughts would be greatly appreciated.
     
  6. Oct 23, 2008 #5

    tiny-tim

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    Hi Mentallic! :smile:

    You're making this far too complicated …

    you don't need the points of contact of the circles …

    just use the radii and the centres …

    if the radii are r1 r2 and r3, and the centres are at C1 C2 and C3, you want the point P (on the line of symmetry) such that PC1 - r1 = PC2 - r2 = PC3 - r3 :smile:

    (and you can use + instead of - for the other solutions :wink:)

    EDIT: btw, why have you called this "Cartesian circles"?

    I just looked that up in wiki, and it's "a mistake in reasoning attributed to René Descartes". :confused:
     
  7. Oct 23, 2008 #6

    Mentallic

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    Thanks tiny-tim, I'll see what results I can gather up from your suggestions when I get the chance.

    I called them Cartesian circles because they are circles on the cartesian number plane :uhh:
    When my classmates saw this question (yes I asked them too) they quickly thought, circle geometry? Anyway I still prefer what I called this to circle geometry, which isn't right either.
     
  8. Oct 24, 2008 #7

    Mentallic

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    Well I took a shot at it, and this is what I got:

    Let the centre of the circle be P(xo,yo)

    [tex]\sqrt{x_o^2+y_o^2}-1=\sqrt{(x_o+1)^2+(y_o-1)^2}-3+2\sqrt{2}=\sqrt{(x_o-\frac{\sqrt{7}}{2})^2+(y_o-\frac{1}{2})^2}-3+2\sqrt{2}[/tex]

    Trying to isolate xo or yo has proven far too difficult for me to accomplish. Are my feudal attempts to find a solution to this problem at a halt, or is the solution to this equation easier than meets the eye?
     
  9. Oct 25, 2008 #8

    tiny-tim

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    ooh, yes, that does look horrible :yuck: …

    (and whyever have you used x0 and y0 instead of x and y?)

    Hint: don't use Cartesian coordinates, go back to good ol' Pythagoras! :biggrin:

    You know that the centre lies on a particular line …

    call its distance along that line from the centre of the big circle r, then use Pythagoras to find its distance to one of the other centres, and make them equal! :smile:
     
  10. Oct 25, 2008 #9

    Mentallic

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    I guess I like to keep coordinates distinguished from variables. My mind might begin to play tricks on me, making me think I'm searching for an equation rather than a coordinate... (I'm sure that lie was a good cover-up :biggrin:)

    ok so I know the point is along [tex]3x+(2-\sqrt{7})y=0[/tex] so using pythagoras, let the distance from the centre to the point [tex]P(x,y)[/tex] be r (notice I'm not using xo :wink:).
    I obtain: [tex]r^2=x^2+y^2[/tex]
    where [tex]3x+(2-\sqrt{7})y=0[/tex]
    therefore [tex]y=(2+\sqrt{7})x[/tex]
    substitute back into distance equation: [tex]r^2=x^2+(x[2+\sqrt{7}])^2[/tex]
    simplified: [tex]r=2x\sqrt{3+\sqrt{7}}[/tex]

    ok so I have an expression for the distance from the centre of the circle needed, to the centre of the unit circle. Tiny-tim, I have a problem with what you have suggested me doing. Finding the distance from the centre of the other circle sounds like a bad idea, simply because the radius of the big circle (unit circle) and one of the others is not the same. It is probably possible to solve the equation in my last post, maybe someone on PF has an idea of how it can be done? Because I know squaring all sides only creates more chaos, I need another approach.

    EDIT:
    oh I just realized what you were asking of me. Yes I could use pythagoras (I guess it could be considered an easier approach to the distance problems) but the end result will still be chaotic like that equation.
    for e.g.
    the distance to the centre of the circle from the outside of the unit circle is given by [tex]\sqrt{x^2+y^2}-1[/tex] however, using the same idea as I did previously in this post, substitute [tex]y=x(2+\sqrt{7})[/tex] since the y coordinate is a function of x, I end up with [tex]r^2=4x^2(3+\sqrt{7})-1[/tex] where if I try to make r the subject, I will just get more chaos again. Because I need to find the distance to the outside of the circles, rather than the centre, this is what creates the difficulty.
     
    Last edited: Oct 25, 2008
  11. Oct 25, 2008 #10

    tiny-tim

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    Hi Mentallic! :smile:

    You're still making this too complicated …

    let the centres be O1 O2 and O3

    let the midpoint of O2 O3 be P, and let O1P meet circle 1 at Q, and let the centre we're looking for be at R

    then you want RQ = RO2 - (√2 - 1), and using Pythagoras that should give you an equation of the form RP + a = √(RP2 + b) :smile:
     
  12. Jan 23, 2009 #11

    gabbagabbahey

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    This is correct and should help you greatly! :smile:

    This equation is incorrect. In fact, you should have 3 separate equations! (actually six- since there are two possible circles ):

    Letting the center of the unknown circle(s) be [itex]P(x,y)[/itex] and letting the Radius of the circle(s) be [itex]R[/itex] you should find that when you calculate the distance from [itex]P[/itex] the centers of each of the other three circles, you get:

    Case 1 the unknown circle actually contains all 3 of the other circles within it:

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

    [tex](x+1)^2+(y-1)^2=\left( R-\sqrt{3-2\sqrt{2}} \right)^2[/tex]

    and

    [tex]\left( x-\frac{\sqrt{7}}{2} \right)^2+\left( y-\frac{1}{2} \right)^2=\left( R-\sqrt{3-2\sqrt{2}} \right)^2[/tex]

    Case 2 the unknown circle doesn't contain any of the other circles within it:

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

    [tex](x+1)^2+(y-1)^2=\left( R+\sqrt{3-2\sqrt{2}} \right)^2[/tex]

    and

    [tex]\left( x-\frac{\sqrt{7}}{2} \right)^2+\left( y-\frac{1}{2} \right)^2=\left( R+\sqrt{3-2\sqrt{2}} \right)^2[/tex]
     
  13. Jan 23, 2009 #12

    Mentallic

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    It looks like you're finding the centre of the circles required by finding the intersection of R satisfying all 3 equations.
    I took a rather backwards approach :tongue2:

    And I have to say, quite ingenious that you subtracted the radius of the circles from the variable radius in the first case :smile:
     
  14. Jan 24, 2009 #13

    Mentallic

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    Isn't this just a system of 3 equations that I need to solve to find a value for R? When I saw this I knew what I needed to do, but doing it has proven much more difficult than expected.

    Since the centre lies on the line:
    [tex]y=(\sqrt{7}+2)x[/tex]

    So I used this fact to simplify this system of equations into 2 equations (I used the first two listed) with 2 variables, x and R.

    [tex]x^2+\left( (\sqrt{7}+2)x \right) ^2=(R-1)^2[/tex]
    and
    [tex](x+1)^2+\left((\sqrt{7}+2)x-1\right)^2=\left( R-\sqrt{3-2\sqrt{2}} \right)^2[/tex]

    I have tried manipulating them in various ways, but I just cannot figure out how I could find R... :confused:
     
  15. Jan 24, 2009 #14

    gabbagabbahey

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    [tex]x^2+\left( (\sqrt{7}+2)x \right) ^2=(R-1)^2[/tex]

    [tex]\implies x^2(1+(\sqrt{7}+2)^2)=4x^2(3+\sqrt{7})=(R-1)^2[/tex]

    [tex]\implies R=2x\sqrt{3+\sqrt{7}}+1[/tex]

    [tex]\implies (x+1)^2+\left((\sqrt{7}+2)x-1\right)^2=\left( 2x\sqrt{3+\sqrt{7}}+1-\sqrt{3-2\sqrt{2}} \right)^2[/tex]

    which gives you a quadratic in x (albeit one with ugly looking coefficients:smile:) which I'm sure you know how to solve.
     
    Last edited: Jan 24, 2009
  16. Jan 25, 2009 #15

    Mentallic

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    I'm not getting the correct answer. Gabbagabbahey did you answer the question too and try graphing it? It would at least be able to support/reject my conclusion that the equations must be incorrect.
     
  17. Jan 25, 2009 #16

    gabbagabbahey

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    Yup, I used these equations and got an answer that worked. I posted a graph in the other thread.
     
  18. Jan 25, 2009 #17

    Hurkyl

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    Here's another approach, if you want to try something interesting.

    There's a transformation called an "inversion"; you pick a point in the plane, a radius p, and then you apply the transformation [itex](r', \theta') = (p/r, \theta)[/itex] (that's polar coordinates centered at your chosen point).

    This transformation maps circles&lines to circles&lines. In particular, if the chosen point is on the intersection of two of your circles, then the inverse of those circes will be a pair of parallel lines. The third circle will still be a circle.

    Now, your problem is to find either:
    (1) A line parallel to the given two and tangent to the circle
    (2) A circle tangent to both lines and to the circle

    A much easier problem. Then, you just repeat the inversion to get your final answer.
     
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