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Math Help - 4 circles and 3 tangents

  1. #1
    Super Member dhiab's Avatar
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    4 circles and 3 tangents

    Find equations for 4 circles each has 3 tangents ,equation of the tangents are :

    \begin{array}{l}<br />
y = - \frac{3}{4}x - 4 \\ <br />
y = \frac{2}{5}x + 4 \\ <br />
y = x + 1 \\ <br />
\end{array}<br />
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  2. #2
    MHF Contributor red_dog's Avatar
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    The three tangents are:

    d_1: 3x+4y+16=0

    d_2: 2x-5y+20=0

    d_3: x-y+1=0

    Let C(a,b) be the center of one circle. Then d(d_1,C)=d(d_2,C)=d(d_3,C)

    \frac{|3a+4b+16|}{5}=\frac{|2a-5b+20|}{\sqrt{29}}=\frac{|a-b+1|}{\sqrt{2}}

    We get four systems:

    \left\{\begin{array}{ll}(3\sqrt{29}-10)a+(4\sqrt{29}+25)+16\sqrt{29}-100=0\\(3\sqrt{2}-5)a+(4\sqrt{2}+5)b+16\sqrt{2}-5=0\end{array}\right.

    \left\{\begin{array}{ll}(3\sqrt{29}-10)a+(4\sqrt{29}+25)+16\sqrt{29}-100=0\\(3\sqrt{2}+5)a+(4\sqrt{2}-5)b+16\sqrt{2}+5=0\end{array}\right.

    \left\{\begin{array}{ll}(3\sqrt{29}+10)a+(4\sqrt{2  9}-25)+16\sqrt{29}+100=0\\(3\sqrt{2}-5)a+(4\sqrt{2}+5)b+16\sqrt{2}-5=0\end{array}\right.

    \left\{\begin{array}{ll}(3\sqrt{29}+10)a+(4\sqrt{2  9}-25)+16\sqrt{29}+100=0\\(3\sqrt{2}+5)a+(4\sqrt{2}-5)b+16\sqrt{2}+5=0\end{array}\right.

    Solving the systems we get the centers of the four circles. Then find their radius and equation.
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