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Math Help - equation of circle

  1. #1
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    Question equation of circle

    I was given this problem last week as a homework.I tried to work it but with avail:
    find the eguation of the circle tangent to the two lines x-3y=-9 and
    3x+y-3=0.thanks for the help I will get
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  2. #2
    MHF Contributor red_dog's Avatar
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    d_1:x-3y+9=0, \ d_2:3x+y-3=0

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

    \frac{|a-3b+9|}{\sqrt{10}}=\frac{|3a+b-3|}{\sqrt{10}}\Rightarrow|a-3b+9|=|3a+b-3|

    1) a-3b+9=3a+b-3\Rightarrow a=6-2b

    Then C(6-2b,b), \ r=\frac{5|3-b|}{\sqrt{10}}

    and the equation is (x+2b-6)^2+(y-b)^2=\frac{5(b-3)^2}{2}, \ b\in\mathbb{R}

    2) a-3b+9=-3a-b+3\Rightarrow b=2a+3

    Then C(a,2a+3), \ r=\frac{5|a|}{\sqrt{10}}

    and the equation is (x-a)^2+(y-2a-3)^2=\frac{5a^2}{2}, \ a\in\mathbb{R}
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