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Math Help - Tangents to circle

  1. #1
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    Tangents to circle

    I can't work out how to answer this question, mental blockage or something!

    Find the values of k such that y=kx are the tangents to the circle x^2+y^2-4x-6y+3=0.

    (Not exactly sure that's the circle I need but I think it should work as it doesn't cross any axes.)

    Thanks!
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  2. #2
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    Quote Originally Posted by Thomas154321 View Post
    I can't work out how to answer this question, mental blockage or something!

    Find the values of k such that y=kx are the tangents to the circle x^2+y^2-4x-6y+3=0.

    (Not exactly sure that's the circle I need but I think it should work as it doesn't cross any axes.)

    Thanks!
    Its a circle
     <br />
(x-2)^2+(y-3)^2={\sqrt{10}}^2<br />
    Its a circle with centre at (2,3) and radius \sqrt{10}
    :::::If you still need help:::
    the line y= kx is tangent to the circle
     <br />
(1+k^2)x^2-(6k+4)x+3=0<br />
    to be tangent there should be only one value of x so
     <br />
(6k+4)^2=12(1+k^2)<br />
    the two values of k are your answer
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