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Math Help - Equation of a Circle

  1. #1
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    Equation of a Circle

    If I have two Circles:
    (x+100)^2 + (y-20)^2 = 20^2
    (x-100)^2 + (y-10)^2 = 10^2
    and a straight line
    y=1

    How can I find the equation of a circle which touch to two circle and tangent to the straight line?

    Thank you!
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  2. #2
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    Quote Originally Posted by mathsrex View Post
    If I have two Circles:
    (x+100)^2 + (y-20)^2 = 20^2
    (x-100)^2 + (y-10)^2 = 10^2
    and a straight line
    y=1

    How can I find the equation of a circle which touch to two circle and tangent to the straight line?

    Thank you!
    The first circle has the center C(-100 , -20) and R_1 = 20. The x-axis is tangent to this circle.

    The second circle has the center C(100 , -10) and R_2 = 10. The x-axis is tangent to this circle.

    Make a rough sketch of the situation. (see attachment. My sketch is not drawn to scale because the distance between the centers of the circles is too large - and the greater circle belongs to the left side)

    You have 2 unknowns: x and \rho. Use the 2 right triangles to set up a system of equations:

    \left| \begin{array}{l}(20+\rho)^2=x^2+(20+1-\rho)^2 \\ (10+\rho)^2=(200-x)^2+(10+1-\rho)^2\end{array} \right.

    You'll get two solutions:

    1. A small circle with it's center at M\left(310 - 10 \cdot \sqrt{861}~,~-\left(\frac{6199}2 - 10 \cdot \sqrt{861} \right) \right) and \rho =  \frac{6201}2 - 100 \cdot \sqrt{861}

    2. A huge circle which touches the 2 circles from below and the tangent point to the straight line is very far away at the right. (I'll only give you the approximate values):

    M(603.4280150, -6033.780150) and r = 6034.780150
    ..
    ..
    Attached Thumbnails Attached Thumbnails Equation of a Circle-tangentialkreis_ankreisgerade.gif  
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  3. #3
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    Thank you very much for your post!

    But I think the center of the two circles should be C(-100 , 20), and C(100 , 10), so the graph may look different. Am I right?
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  4. #4
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    Thank you very much for your post!

    But I think the center of the two circles should be C(-100 , 20), and C(100 , 10), so the graph may look different. Am I right?
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  5. #5
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    Quote Originally Posted by mathsrex View Post
    Thank you very much for your post!

    But I think the center of the two circles should be C(-100 , 20), and C(100 , 10), so the graph may look different. Am I right?
    You're right, I guess earboth misread it
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  6. #6
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    Quote Originally Posted by wingless View Post
    You're right, I guess earboth misread it
    Yes, but only a little bit.
    1. It is quite difficult to place the centers at (-100, 20) and (100, 10). Therefore I took as an example (-10, -10) and (10, - 20)

    2. Of course, you are right, the circles should be above the x-axis.

    3. All my considerations should be OK. Therefore I will not post a repaired version.
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  7. #7
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    Quote Originally Posted by mathsrex View Post
    If I have two Circles:
    (x+100)^2 + (y-20)^2 = 20^2
    (x-100)^2 + (y-10)^2 = 10^2
    and a straight line
    y=1

    How can I find the equation of a circle which touch to two circle and tangent to the straight line?

    Thank you!
    Draw the figure so that you can follow.
    There is a large circle of radius 20 units that is centered at point A(100,20). Then there is a small circle of radius 10 units that is centered at point B(100,10). Then there is that horizontal line y=1.

    The circle we are looking for, P, that is tangent to all three....large and small circles and y=1 line... should be centered somehere in the 1st quadrant because of the location of the small circle B.
    Let's call the center of circle P as point P(u,v). Its radius, we call r.

    The tangent line from P to the horizontal line y=1 is a vertical line segment. This vertical tangent line is r.
    So, r = v -1 --------(i)

    Draw a line connecting the P to the center of circle A. This line segment is (r +20)
    Draw a horizontal line connecting centerpoint A to the vertical r at point C. The length of this AC line segment is (100 +u). And, the length of PC is (v -20)
    So in the right triangle PCA,
    hypotenuse = r +20 = (v -1) +20 = v +19
    vertical leg = v -20
    horizontal leg = 100 +u
    By Pythagorean Theorem,
    (v +19)^2 = (v -20)^2 +(100 +u)^2 -----------(ii)

    Draw a line connecting the P to the center of circle B. This line segment is (r +10)
    Draw a horizontal line connecting centerpoint B to the vertical r at point D. The length of this BD line segment is (100 -u)
    And, the length of PD is (v -10)
    So in the right triangle PDB,
    hypotenuse = r +10 = (v -1) +10 = v +9
    vertical leg = v -10
    horizontal leg = 100 -u
    By Pythagorean Theorem,
    (v +9)^2 = (v -10)^2 +(100 -u)^2 -----------(iii)

    Now solve Eq.(ii) and Eq.(iii) simultaneously to find u and v.

    From v you can get r.

    So, the circle P in question is
    (x -u)^2 +(y -v)^2 = r^2

    ((I'm getting u = 17.78685; v = 178.36849; r = 177.36849))
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  8. #8
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    I did my calculations again:

    According to ticbol you are looking for a circle with C(p, q) and radius r.

    Then you can set up a system of three equations:

    \left|\begin{array}{l}(r+20)^2 = (p+100)^2+(q-20)^2 \\ (r+10)^2 = (p-100)^2 + (q-10)^2 \\ r = q-1 \end{array}\right.

    This system will yield two solutions:

    1. (ticbols solution)
    C\left(290-10 \cdot \sqrt{741}~,~ \frac{5801}2 - 100\cdot \sqrt{741}\right), r = \frac{5799}2 - 100\cdot \sqrt{741} Approximately:

    C( 17.787, 178.368), r = 177.368

    2.

    C\left(290+10 \cdot \sqrt{741}~,~ \frac{5801}2 + 100\cdot \sqrt{741}\right), r = \frac{5799}2 + 100\cdot \sqrt{741} Approximately:

    C( 562.213, 5622.632), r = 5621.632

    EDIT: I've attached a sketch of the situation. The red line is a small part of the big circle of solution #2.
    Attached Thumbnails Attached Thumbnails Equation of a Circle-tang_an2krse.gif  
    Last edited by earboth; August 6th 2008 at 10:15 AM.
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  9. #9
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    Since both circles cut the line y = 1 there must be at least one more circles which touches the circles and the line from below:

    \left|\begin{array}{l}(r+20)^2 = (p+100)^2+(q-20)^2 \\ (r+10)^2 = (p-100)^2 + (q-10)^2 \\ r = -q+1 \end{array}\right.

    This system of equations has only one solution:

    C([ 0.05, -4984.50), r = 4985.50

    I've attached a sketch of the circles. The green line is a part of the last circle, touching the small circles and the line from below.
    Attached Thumbnails Attached Thumbnails Equation of a Circle-tangentialkreis2_ankreisgerade.gif  
    Last edited by earboth; August 6th 2008 at 11:49 AM.
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