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Math Help - coordinate geometry

  1. #1
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    coordinate geometry

    find the equation of circle passing through the points of contact of direct common tangents of x^2+y^2=16 and x^2+y^2-12x+32=0
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  2. #2
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    Quote Originally Posted by prasum View Post
    find the equation of circle passing through the points of contact of direct common tangents of x^2+y^2=16 and x^2+y^2-12x+32=0
    1. The circle c_1: x^2+y^2=4^2 has the center (0, 0) and the radius 4.
    The circle c_2:x^2+y^2-12x+32=0~\implies~(x-6)^2+y^2=2^2 has the center (6, 0) and the radius 2.

    2. Use similar right triangles to determine the missing lengthes. I used Euklid's theorem to calculate the coordinates of the tangent points: T_1\left(\frac43\ ,\ \frac83 \sqrt{2}\right)

    and T_2\left(\frac{20}3\ ,\ \frac43 \sqrt{2}\right)

    3. The center of the resulting circle c_r must be on the x-axis and on the perpendicular bisector of \overline{T_1T_2}
    The slope of this perpendicular bisector equals the slope of \overline{OT_1} = \dfrac{\frac83 \sqrt{2}}{\frac43}=2\sqrt{2}

    The midpoint of \overline{T_1T_2} has the coordinates M\left(\dfrac{\frac43 + \frac{20}3}2\ , \ \dfrac{\frac83 \sqrt{2} + \frac{4}3 \sqrt{2}}2 \right). That means M(4, 2\sqrt{2})

    4. The perpendicular bisector through M has the equation:

    y-2 \sqrt{2} = 2\sqrt{2}(x-4)

    This line crosses the x-axis at:

    -2 \sqrt{2} = 2\sqrt{2}(x-4)~\implies~-1=x-4~\implies~x=3

    5. The center of the resulting circle is C(3, 0). Now calculate the distance |\overline{CT_1}| to get the radius of c_r. I've got r=\sqrt{17}. Therefore the equation of c_r is:

    c_r: (x-3)^2+y^2=17
    Attached Thumbnails Attached Thumbnails coordinate geometry-gemeinskrs_4pkte.png  
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  3. #3
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    Thanks earboth a nice solution

    bjh
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    thanks man
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  5. #5
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    how did yu use euklids geometry here and what is it basically
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  6. #6
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    Quote Originally Posted by prasum View Post
    how did yu use euklids geometry here and what is it basically
    1. I learned that in English Euclid's theorem is related to number theory. In Germany a group of theorems is summed under the name Euclid's theorems where this theorem belongs to:

    In a right triangle the height divides the hypotenuse into two segments, p and q. Then

    h^2 = p \cdot q

    And that is what I used:

    2. Take c_1 and the corresponding right triangle with

    c = 12, b = 4, a = 8 \sqrt{8}

    According to the the attached sketch the tangent point T_1 has the coordinates T_1(x, h)

    3. From the area of the right tringle you know:

    A=\frac12 \cdot c \cdot h = \frac12 \cdot a \cdot b ~\implies~h = \dfrac{a \cdot b}{c}

    4. With y = 12 - x you get:

    x(12-x)=h^2

    Since you know already the value of h you can solve the equation for x.

    5. Alternatively you can plug in the value of h instead of y into the equation of the circle to get the x-coordinate of the tangent point:

    x^2+\dfrac{16 \cdot 128}{144} = 16~\implies~ |x|=\dfrac34
    Attached Thumbnails Attached Thumbnails coordinate geometry-euklid_theorem2.png  
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