If the line $lx+my=1$ touches the circle $x^2+y^2=a^2$, prove












1












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If the line $lx+my=1$ touches the circle $x^2+y^2=a^2$, prove that the point $(l,m)$ lies on a circle with radius $dfrac{1}{a}$.




My Attempt:



The center and radius of the circle $x^2+y^2=a^2$ are $(0,0)$ and $a$, respectively. Since $lx+my-1=0$ touches the circle,
$$a=left|dfrac {lcdot0+mcdot0-1}{sqrt {l^2+m^2}}right| implies a^2=dfrac {1}{l^2+m^2}.$$






analytic-geometry circle







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  • $begingroup$
    What’s your question?
    $endgroup$
    – amd
    Dec 25 '17 at 1:22










  • $begingroup$
    @egreg, What? Could you please explain?
    $endgroup$
    – blue_eyed_...
    Dec 25 '17 at 1:26
















1












$begingroup$



If the line $lx+my=1$ touches the circle $x^2+y^2=a^2$, prove that the point $(l,m)$ lies on a circle with radius $dfrac{1}{a}$.




My Attempt:



The center and radius of the circle $x^2+y^2=a^2$ are $(0,0)$ and $a$, respectively. Since $lx+my-1=0$ touches the circle,
$$a=left|dfrac {lcdot0+mcdot0-1}{sqrt {l^2+m^2}}right| implies a^2=dfrac {1}{l^2+m^2}.$$






analytic-geometry circle







share|cite|improve this question











$endgroup$












  • $begingroup$
    What’s your question?
    $endgroup$
    – amd
    Dec 25 '17 at 1:22










  • $begingroup$
    @egreg, What? Could you please explain?
    $endgroup$
    – blue_eyed_...
    Dec 25 '17 at 1:26














1












1








1





$begingroup$



If the line $lx+my=1$ touches the circle $x^2+y^2=a^2$, prove that the point $(l,m)$ lies on a circle with radius $dfrac{1}{a}$.




My Attempt:



The center and radius of the circle $x^2+y^2=a^2$ are $(0,0)$ and $a$, respectively. Since $lx+my-1=0$ touches the circle,
$$a=left|dfrac {lcdot0+mcdot0-1}{sqrt {l^2+m^2}}right| implies a^2=dfrac {1}{l^2+m^2}.$$






analytic-geometry circle







share|cite|improve this question











$endgroup$





If the line $lx+my=1$ touches the circle $x^2+y^2=a^2$, prove that the point $(l,m)$ lies on a circle with radius $dfrac{1}{a}$.




My Attempt:



The center and radius of the circle $x^2+y^2=a^2$ are $(0,0)$ and $a$, respectively. Since $lx+my-1=0$ touches the circle,
$$a=left|dfrac {lcdot0+mcdot0-1}{sqrt {l^2+m^2}}right| implies a^2=dfrac {1}{l^2+m^2}.$$






analytic-geometry circle




analytic-geometry circle






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '17 at 1:57









Math Lover

14k31436




14k31436










asked Dec 25 '17 at 1:17









blue_eyed_...blue_eyed_...

3,25921649




3,25921649












  • $begingroup$
    What’s your question?
    $endgroup$
    – amd
    Dec 25 '17 at 1:22










  • $begingroup$
    @egreg, What? Could you please explain?
    $endgroup$
    – blue_eyed_...
    Dec 25 '17 at 1:26


















  • $begingroup$
    What’s your question?
    $endgroup$
    – amd
    Dec 25 '17 at 1:22










  • $begingroup$
    @egreg, What? Could you please explain?
    $endgroup$
    – blue_eyed_...
    Dec 25 '17 at 1:26
















$begingroup$
What’s your question?
$endgroup$
– amd
Dec 25 '17 at 1:22




$begingroup$
What’s your question?
$endgroup$
– amd
Dec 25 '17 at 1:22












$begingroup$
@egreg, What? Could you please explain?
$endgroup$
– blue_eyed_...
Dec 25 '17 at 1:26




$begingroup$
@egreg, What? Could you please explain?
$endgroup$
– blue_eyed_...
Dec 25 '17 at 1:26










1 Answer
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1












$begingroup$

That's perfectly right if you are allowed to use synthetic geometry results, namely that the tangent is orthogonal to the radius at the point of contact, so the distance of the line from the center of the circle equals the radius.



If the proof must be purely analytic, it's a bit more complicated.



I suggest to write points on the line $lx+my=1$ in the form
$$
x=-mt, qquad y=lt+1/m
$$
(the case where $m=0$ can be treated as a special case). The intersections with the given circle can be computed from
$$
m^2t^2+l^2t^2+2frac{l}{m}t+frac{1}{m^2}-a^2=0
$$
and the discriminant of the polynomial must be $0$, so
$$
frac{l^2}{m^2}-(m^2+l^2)left(frac{1}{m^2}-a^2right)=0
$$
which reduces to
$$
-1+a^2(m^2+l^2)=0
$$
which is the thesis.



For $m=0$ the line is $lx=1$ and the intersection with the circle are
$$
frac{1}{l^2}+y^2=a^2
$$
Since this equation must be satisfied by a single point, we get $l^2=frac{1}{a^2}$.






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    1 Answer
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    active

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    1 Answer
    1






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    active

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    active

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    1












    $begingroup$

    That's perfectly right if you are allowed to use synthetic geometry results, namely that the tangent is orthogonal to the radius at the point of contact, so the distance of the line from the center of the circle equals the radius.



    If the proof must be purely analytic, it's a bit more complicated.



    I suggest to write points on the line $lx+my=1$ in the form
    $$
    x=-mt, qquad y=lt+1/m
    $$
    (the case where $m=0$ can be treated as a special case). The intersections with the given circle can be computed from
    $$
    m^2t^2+l^2t^2+2frac{l}{m}t+frac{1}{m^2}-a^2=0
    $$
    and the discriminant of the polynomial must be $0$, so
    $$
    frac{l^2}{m^2}-(m^2+l^2)left(frac{1}{m^2}-a^2right)=0
    $$
    which reduces to
    $$
    -1+a^2(m^2+l^2)=0
    $$
    which is the thesis.



    For $m=0$ the line is $lx=1$ and the intersection with the circle are
    $$
    frac{1}{l^2}+y^2=a^2
    $$
    Since this equation must be satisfied by a single point, we get $l^2=frac{1}{a^2}$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      That's perfectly right if you are allowed to use synthetic geometry results, namely that the tangent is orthogonal to the radius at the point of contact, so the distance of the line from the center of the circle equals the radius.



      If the proof must be purely analytic, it's a bit more complicated.



      I suggest to write points on the line $lx+my=1$ in the form
      $$
      x=-mt, qquad y=lt+1/m
      $$
      (the case where $m=0$ can be treated as a special case). The intersections with the given circle can be computed from
      $$
      m^2t^2+l^2t^2+2frac{l}{m}t+frac{1}{m^2}-a^2=0
      $$
      and the discriminant of the polynomial must be $0$, so
      $$
      frac{l^2}{m^2}-(m^2+l^2)left(frac{1}{m^2}-a^2right)=0
      $$
      which reduces to
      $$
      -1+a^2(m^2+l^2)=0
      $$
      which is the thesis.



      For $m=0$ the line is $lx=1$ and the intersection with the circle are
      $$
      frac{1}{l^2}+y^2=a^2
      $$
      Since this equation must be satisfied by a single point, we get $l^2=frac{1}{a^2}$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        That's perfectly right if you are allowed to use synthetic geometry results, namely that the tangent is orthogonal to the radius at the point of contact, so the distance of the line from the center of the circle equals the radius.



        If the proof must be purely analytic, it's a bit more complicated.



        I suggest to write points on the line $lx+my=1$ in the form
        $$
        x=-mt, qquad y=lt+1/m
        $$
        (the case where $m=0$ can be treated as a special case). The intersections with the given circle can be computed from
        $$
        m^2t^2+l^2t^2+2frac{l}{m}t+frac{1}{m^2}-a^2=0
        $$
        and the discriminant of the polynomial must be $0$, so
        $$
        frac{l^2}{m^2}-(m^2+l^2)left(frac{1}{m^2}-a^2right)=0
        $$
        which reduces to
        $$
        -1+a^2(m^2+l^2)=0
        $$
        which is the thesis.



        For $m=0$ the line is $lx=1$ and the intersection with the circle are
        $$
        frac{1}{l^2}+y^2=a^2
        $$
        Since this equation must be satisfied by a single point, we get $l^2=frac{1}{a^2}$.






        share|cite|improve this answer









        $endgroup$



        That's perfectly right if you are allowed to use synthetic geometry results, namely that the tangent is orthogonal to the radius at the point of contact, so the distance of the line from the center of the circle equals the radius.



        If the proof must be purely analytic, it's a bit more complicated.



        I suggest to write points on the line $lx+my=1$ in the form
        $$
        x=-mt, qquad y=lt+1/m
        $$
        (the case where $m=0$ can be treated as a special case). The intersections with the given circle can be computed from
        $$
        m^2t^2+l^2t^2+2frac{l}{m}t+frac{1}{m^2}-a^2=0
        $$
        and the discriminant of the polynomial must be $0$, so
        $$
        frac{l^2}{m^2}-(m^2+l^2)left(frac{1}{m^2}-a^2right)=0
        $$
        which reduces to
        $$
        -1+a^2(m^2+l^2)=0
        $$
        which is the thesis.



        For $m=0$ the line is $lx=1$ and the intersection with the circle are
        $$
        frac{1}{l^2}+y^2=a^2
        $$
        Since this equation must be satisfied by a single point, we get $l^2=frac{1}{a^2}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 25 '17 at 1:32









        egregegreg

        181k1485203




        181k1485203






























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