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Math Help - circle-tangents question

  1. #1
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    Sep 2005
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    plz assist with circle-tangents question

    I really need some help on this problem:

    1. The line x-2y=-4 is tangent to a circle at (0,2). The line y=2x-7 is tangent to the same circle at (3,-1). Find the center of the circle.

    *If the equation of the circle is x^2 + y^2 = r^2 and the equation of the tangent line is y=mx+b, show that-
    (a) r^2(1+m^2)= b^2
    hint: the quadratic equation x^2 + (mx+b)^2 has exactly one solution
    (b) the point of tangency is (-r^2 * m/b, r^2/b)
    (c) the tangent line is perpendicular to the line containing the center of the circle and the point of tangency
    Last edited by MathGuru; September 11th 2005 at 05:37 PM.
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  2. #2
    Junior Member
    Joined
    Aug 2005
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    53

    Circle

    Hi,

    Ok, the main concept is to see that having a line tangent to a circle at a point P means that the radius passing by the point P is perpendicular to that line at that point.

    Numerically, the line x-2y=-4 (same as y= x/2 + 2) is tangent to the circle at (0;2). Let's draw the radius going from the center of the circle to (0;2). We know that the line L1 containing this radius is perpendicular to y= x/2 + 2. So you can use the property that when two lines are perpendicular, the slope of the first line is equal to -1/(the slope of the second line). Cool huh ?

    So the line y= x/2 + 2 has slope 1/2 so the slope of L1 is -1/(1/2) = -2. So the equation of L1 should be y1=-2x+b. And we know that is passes by (0;2) because it intesects the first line at that point so we put it in the equation :

    y1=-2x+b so 2=-2 (0) + b so b=2. Great ! Now, y1=-2x+2.

    Cool, we know that the center is somewhere on the line y1 but where ?
    So it is here that we need the seconde tangent y=2x-7.

    We use the same trick to find the second line L2 which contains the radius passing by (3;-1). Since it is perpendicular to y=2x-7, the slope of L2 is -1/(2) = -1/2. So again the equation of L2 is y2=-1/2 x + B. And we know that is passes by (3;-1) so

    y2=-1/2 x + B so -1=-1/2 (3) + B so B=1/2. Ok y2=-1/2x+1/2.

    So we know that the center of the circle (x,y) is the interection of L1 and L2 so at that point the two lines have the same x and the same y so we put
    y1=y2
    so -2x+2 = -1/2 x+1/2.
    We multiply both sides by 2 : -4x+4 = -x+1.
    We put the x together and the other numbers together : -3x=-3 so x = 1.
    Now we find the y of the center by replacing the x=1 into one of the two equation y1 or y2 since they have the same y at that point it should be equal :
    y1 = -2(1)+2 = 0 and y2=-1/2 (1)+1/2=0 Great ! Done

    So the center is (1;0)
    Attached Thumbnails Attached Thumbnails circle-tangents question-cercle-tg.gif  
    Last edited by hemza; September 12th 2005 at 08:30 AM.
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  3. #3
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    Sep 2005
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    Thumbs up Wow, thanks

    Thanks, now i understand it due to your thorough explaination.
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