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Math Help - length of AB

  1. #1
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    length of AB

    the line y = mx is a tangent to the circle x^2 + y^2 - 10y + 16 = 0
    a) find the two possible values of m
    b) the tangents meet the circle at points A and B. find the length of AB

    i've done a) and its definitely m = -\frac {4}{3} and \frac {4}{3}

    but i don't know how to tackle b), could someone show me please? thankyou
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  2. #2
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    Quote Originally Posted by mark View Post
    the line y = mx is a tangent to the circle x^2 + y^2 - 10y + 16 = 0
    a) find the two possible values of m
    b) the tangents meet the circle at points A and B. find the length of AB

    i've done a) and its definitely m = -\frac {4}{3} and \frac {4}{3}

    but i don't know how to tackle b), could someone show me please? thankyou
    I assume that you have calculated the coordinates of the points of intersection. Then you've got:

    x=\dfrac{5m \pm\sqrt{9m^2-16}}{1+m^2}

    Plug in the values for m to get the x-coordinate of A or B. Plug in this x-value into the equation of the tangent to get the y-coordinate. Use the distance formula to calculate the length.
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  3. #3
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    Well, the obvious way would be to find the points A and B and use the distance formula! And that's easy.

    I presume that you found 4/3 and -4/3 by replacing y in [tex]x^2+ y^2- 10y+ 16= 0[tex] by mx to get (m^2+ 1)x^2- 10mx+ 16= 0 and arguing that, since a tangent line touches the circle in only one place, that equation must have a single solution for x and so it discriminant, b^2- 4ac= (-10m)^2- 4(16)(1+ m^2)= 0. Solving that for m gave you the two solutions.

    And, since the discrimant is 0, the quadratic formula reduces to x= -\frac{b}{2a}= \frac{5m}{1+ m^2}. Put m equal to 4/3 and -4/3 in that to find the two values of x. Of course, y is just mx.

    And, in fact, I note that the center of the circle is on the y-axis so the y values of points A and B are the same! The distance from A to B is just the difference of the x coordinates. How simple is that?
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  4. #4
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    ok i've used the simplified quadratic formula (because the discriminant is 0) \frac{5m}{1+ m^2} then \frac{5(4/3)}{1 + (16/9)} which came to \frac {60/9}{25/9} and i've arrived at x = 2.4 and y = 3.2.
    then using the distance formula like so \sqrt {2.4^2 + 3.2^2} which come to \sqrt {5.76 + 10.24} which equals \sqrt {16} which is 4. the answer given to me by the book is 4.8 though? what have i done wrong here?
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  5. #5
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    Quote Originally Posted by mark View Post
    ok i've used the simplified quadratic formula (because the discriminant is 0) \frac{5m}{1+ m^2} then \frac{5(4/3)}{1 + (16/9)} which came to \frac {60/9}{25/9} and i've arrived at x = 2.4 and y = 3.2. <<<<<<<<< That means the point A is A(2.4, 3.2)

    Now use \color{blue}\bold{m=-\dfrac43} to get the coordinates of B(-2.4, 3.2)
    then using the distance formula like so \sqrt {2.4^2 + 3.2^2} which come to \sqrt {5.76 + 10.24} which equals \sqrt {16} which is 4. the answer given to me by the book is 4.8 though? what have i done wrong here?
    Please read HallsofIvy's post again. He's already given the solution.
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  6. #6
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    Quote Originally Posted by mark View Post
    ok i've used the simplified quadratic formula (because the discriminant is 0) \frac{5m}{1+ m^2} then \frac{5(4/3)}{1 + (16/9)} which came to \frac {60/9}{25/9} and i've arrived at x = 2.4 and y = 3.2.
    then using the distance formula like so \sqrt {2.4^2 + 3.2^2} which come to \sqrt {5.76 + 10.24} which equals \sqrt {16} which is 4. the answer given to me by the book is 4.8 though? what have i done wrong here?
    they are on the same y-axis!
    draw it out if you still don't understand!
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