How to find a tangent to a circle from an external point using calculus?












0












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So I know how to find the tangent from an external point using algebra but that involves many equations making the entire process tedious. Anyways I have a calculus exam coming up and I think I should be using calculus to solve such problems .






calculus geometry







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  • $begingroup$
    How many equations do you consider “many?” If you can write down the equation of the circle, you can also write down the equation of the polar of a point. The points of tangency are the solutions to this system of two equations. That doesn’t sound particularly tedious to me.
    $endgroup$
    – amd
    Jan 12 at 9:32












  • $begingroup$
    Why do we need calculus to begin with? Analytic geometry works perfectly well.
    $endgroup$
    – Oscar Lanzi
    Jan 12 at 11:11
















0












$begingroup$


So I know how to find the tangent from an external point using algebra but that involves many equations making the entire process tedious. Anyways I have a calculus exam coming up and I think I should be using calculus to solve such problems .






calculus geometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    How many equations do you consider “many?” If you can write down the equation of the circle, you can also write down the equation of the polar of a point. The points of tangency are the solutions to this system of two equations. That doesn’t sound particularly tedious to me.
    $endgroup$
    – amd
    Jan 12 at 9:32












  • $begingroup$
    Why do we need calculus to begin with? Analytic geometry works perfectly well.
    $endgroup$
    – Oscar Lanzi
    Jan 12 at 11:11














0












0








0


0



$begingroup$


So I know how to find the tangent from an external point using algebra but that involves many equations making the entire process tedious. Anyways I have a calculus exam coming up and I think I should be using calculus to solve such problems .






calculus geometry







share|cite|improve this question









$endgroup$




So I know how to find the tangent from an external point using algebra but that involves many equations making the entire process tedious. Anyways I have a calculus exam coming up and I think I should be using calculus to solve such problems .






calculus geometry




calculus geometry






share|cite|improve this question













share|cite|improve this question











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asked Jan 12 at 6:34









Ishaan ParikhIshaan Parikh

256




256












  • $begingroup$
    How many equations do you consider “many?” If you can write down the equation of the circle, you can also write down the equation of the polar of a point. The points of tangency are the solutions to this system of two equations. That doesn’t sound particularly tedious to me.
    $endgroup$
    – amd
    Jan 12 at 9:32












  • $begingroup$
    Why do we need calculus to begin with? Analytic geometry works perfectly well.
    $endgroup$
    – Oscar Lanzi
    Jan 12 at 11:11


















  • $begingroup$
    How many equations do you consider “many?” If you can write down the equation of the circle, you can also write down the equation of the polar of a point. The points of tangency are the solutions to this system of two equations. That doesn’t sound particularly tedious to me.
    $endgroup$
    – amd
    Jan 12 at 9:32












  • $begingroup$
    Why do we need calculus to begin with? Analytic geometry works perfectly well.
    $endgroup$
    – Oscar Lanzi
    Jan 12 at 11:11
















$begingroup$
How many equations do you consider “many?” If you can write down the equation of the circle, you can also write down the equation of the polar of a point. The points of tangency are the solutions to this system of two equations. That doesn’t sound particularly tedious to me.
$endgroup$
– amd
Jan 12 at 9:32






$begingroup$
How many equations do you consider “many?” If you can write down the equation of the circle, you can also write down the equation of the polar of a point. The points of tangency are the solutions to this system of two equations. That doesn’t sound particularly tedious to me.
$endgroup$
– amd
Jan 12 at 9:32














$begingroup$
Why do we need calculus to begin with? Analytic geometry works perfectly well.
$endgroup$
– Oscar Lanzi
Jan 12 at 11:11




$begingroup$
Why do we need calculus to begin with? Analytic geometry works perfectly well.
$endgroup$
– Oscar Lanzi
Jan 12 at 11:11










3 Answers
3






active

oldest

votes


















0












$begingroup$

Equation of a circle is degree 2. Equation of a line passing through a given point is degree 1. So their intersection is obtained by solving a quadratic equation. For a tangent, they have only one point in commun, so the discriminant is 0...






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The OP asked for a solution that uses calculus.
    $endgroup$
    – amd
    Jan 12 at 9:23










  • $begingroup$
    No. The OP said he thought he should be using calculus. In fact you don't. This approach is a good one but I think it should be developed further anyway, to be useful.
    $endgroup$
    – Oscar Lanzi
    Jan 12 at 11:15



















0












$begingroup$

We maximize slope from a given fixed point $ (x_1,y_1)$ to a parameterized circle variable point in its standard form:



$$ (x-h) = a cos t,, (y-k)= a sin t; tag1 $$
and solve for the parameter



$$ frac{y-y_1}{x-x_1}= frac{k+ a sin t -y_1}{h+ a cos t -x_1}=frac{-cos t} {sin t}tag2$$



the latter has been obtained by chain rule differentiation



Simplifying get the condition to obtain $t_1$ of tangent points (two)



$$ sin t_1 (k-y_1)+ cos t_1 (h-x_1)+a=0 tag3 $$



It can be also recognized as tangent-normal form of a straight line.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Let us call $R$ the ray and $C = (x_0, y_0)$ the center of the circle. The points of the circle are the $M =(Rcostheta, Rsintheta)$.

    Let $E = (a, b)$ be the external point.

    Then the two points $M$ corresponding to a tangeant passing through $E$ are such that vector $vec{CM}$ is orthogonal to $vec{EM}$.



    It immediately gives:
    $$costheta,(x_0+ Rcostheta - a)+ sintheta,(y_0+Rsintheta-b) = 0$$
    And then
    $$costheta,(x_0-a) + sintheta,(y_0-b) + R$$



    One possibility to go further is to rewrite $vec{EC}$ as $(rhocosphi, rhosinphi )$

    And therefore
    $$cos(theta - phi) = -frac{R}{rho}$$






    share|cite|improve this answer











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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      Equation of a circle is degree 2. Equation of a line passing through a given point is degree 1. So their intersection is obtained by solving a quadratic equation. For a tangent, they have only one point in commun, so the discriminant is 0...






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        The OP asked for a solution that uses calculus.
        $endgroup$
        – amd
        Jan 12 at 9:23










      • $begingroup$
        No. The OP said he thought he should be using calculus. In fact you don't. This approach is a good one but I think it should be developed further anyway, to be useful.
        $endgroup$
        – Oscar Lanzi
        Jan 12 at 11:15
















      0












      $begingroup$

      Equation of a circle is degree 2. Equation of a line passing through a given point is degree 1. So their intersection is obtained by solving a quadratic equation. For a tangent, they have only one point in commun, so the discriminant is 0...






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        The OP asked for a solution that uses calculus.
        $endgroup$
        – amd
        Jan 12 at 9:23










      • $begingroup$
        No. The OP said he thought he should be using calculus. In fact you don't. This approach is a good one but I think it should be developed further anyway, to be useful.
        $endgroup$
        – Oscar Lanzi
        Jan 12 at 11:15














      0












      0








      0





      $begingroup$

      Equation of a circle is degree 2. Equation of a line passing through a given point is degree 1. So their intersection is obtained by solving a quadratic equation. For a tangent, they have only one point in commun, so the discriminant is 0...






      share|cite|improve this answer









      $endgroup$



      Equation of a circle is degree 2. Equation of a line passing through a given point is degree 1. So their intersection is obtained by solving a quadratic equation. For a tangent, they have only one point in commun, so the discriminant is 0...







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 12 at 6:49









      amaama

      335137




      335137












      • $begingroup$
        The OP asked for a solution that uses calculus.
        $endgroup$
        – amd
        Jan 12 at 9:23










      • $begingroup$
        No. The OP said he thought he should be using calculus. In fact you don't. This approach is a good one but I think it should be developed further anyway, to be useful.
        $endgroup$
        – Oscar Lanzi
        Jan 12 at 11:15


















      • $begingroup$
        The OP asked for a solution that uses calculus.
        $endgroup$
        – amd
        Jan 12 at 9:23










      • $begingroup$
        No. The OP said he thought he should be using calculus. In fact you don't. This approach is a good one but I think it should be developed further anyway, to be useful.
        $endgroup$
        – Oscar Lanzi
        Jan 12 at 11:15
















      $begingroup$
      The OP asked for a solution that uses calculus.
      $endgroup$
      – amd
      Jan 12 at 9:23




      $begingroup$
      The OP asked for a solution that uses calculus.
      $endgroup$
      – amd
      Jan 12 at 9:23












      $begingroup$
      No. The OP said he thought he should be using calculus. In fact you don't. This approach is a good one but I think it should be developed further anyway, to be useful.
      $endgroup$
      – Oscar Lanzi
      Jan 12 at 11:15




      $begingroup$
      No. The OP said he thought he should be using calculus. In fact you don't. This approach is a good one but I think it should be developed further anyway, to be useful.
      $endgroup$
      – Oscar Lanzi
      Jan 12 at 11:15











      0












      $begingroup$

      We maximize slope from a given fixed point $ (x_1,y_1)$ to a parameterized circle variable point in its standard form:



      $$ (x-h) = a cos t,, (y-k)= a sin t; tag1 $$
      and solve for the parameter



      $$ frac{y-y_1}{x-x_1}= frac{k+ a sin t -y_1}{h+ a cos t -x_1}=frac{-cos t} {sin t}tag2$$



      the latter has been obtained by chain rule differentiation



      Simplifying get the condition to obtain $t_1$ of tangent points (two)



      $$ sin t_1 (k-y_1)+ cos t_1 (h-x_1)+a=0 tag3 $$



      It can be also recognized as tangent-normal form of a straight line.






      share|cite|improve this answer











      $endgroup$


















        0












        $begingroup$

        We maximize slope from a given fixed point $ (x_1,y_1)$ to a parameterized circle variable point in its standard form:



        $$ (x-h) = a cos t,, (y-k)= a sin t; tag1 $$
        and solve for the parameter



        $$ frac{y-y_1}{x-x_1}= frac{k+ a sin t -y_1}{h+ a cos t -x_1}=frac{-cos t} {sin t}tag2$$



        the latter has been obtained by chain rule differentiation



        Simplifying get the condition to obtain $t_1$ of tangent points (two)



        $$ sin t_1 (k-y_1)+ cos t_1 (h-x_1)+a=0 tag3 $$



        It can be also recognized as tangent-normal form of a straight line.






        share|cite|improve this answer











        $endgroup$
















          0












          0








          0





          $begingroup$

          We maximize slope from a given fixed point $ (x_1,y_1)$ to a parameterized circle variable point in its standard form:



          $$ (x-h) = a cos t,, (y-k)= a sin t; tag1 $$
          and solve for the parameter



          $$ frac{y-y_1}{x-x_1}= frac{k+ a sin t -y_1}{h+ a cos t -x_1}=frac{-cos t} {sin t}tag2$$



          the latter has been obtained by chain rule differentiation



          Simplifying get the condition to obtain $t_1$ of tangent points (two)



          $$ sin t_1 (k-y_1)+ cos t_1 (h-x_1)+a=0 tag3 $$



          It can be also recognized as tangent-normal form of a straight line.






          share|cite|improve this answer











          $endgroup$



          We maximize slope from a given fixed point $ (x_1,y_1)$ to a parameterized circle variable point in its standard form:



          $$ (x-h) = a cos t,, (y-k)= a sin t; tag1 $$
          and solve for the parameter



          $$ frac{y-y_1}{x-x_1}= frac{k+ a sin t -y_1}{h+ a cos t -x_1}=frac{-cos t} {sin t}tag2$$



          the latter has been obtained by chain rule differentiation



          Simplifying get the condition to obtain $t_1$ of tangent points (two)



          $$ sin t_1 (k-y_1)+ cos t_1 (h-x_1)+a=0 tag3 $$



          It can be also recognized as tangent-normal form of a straight line.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 12 at 8:05

























          answered Jan 12 at 7:56









          NarasimhamNarasimham

          21.2k62258




          21.2k62258























              0












              $begingroup$

              Let us call $R$ the ray and $C = (x_0, y_0)$ the center of the circle. The points of the circle are the $M =(Rcostheta, Rsintheta)$.

              Let $E = (a, b)$ be the external point.

              Then the two points $M$ corresponding to a tangeant passing through $E$ are such that vector $vec{CM}$ is orthogonal to $vec{EM}$.



              It immediately gives:
              $$costheta,(x_0+ Rcostheta - a)+ sintheta,(y_0+Rsintheta-b) = 0$$
              And then
              $$costheta,(x_0-a) + sintheta,(y_0-b) + R$$



              One possibility to go further is to rewrite $vec{EC}$ as $(rhocosphi, rhosinphi )$

              And therefore
              $$cos(theta - phi) = -frac{R}{rho}$$






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                Let us call $R$ the ray and $C = (x_0, y_0)$ the center of the circle. The points of the circle are the $M =(Rcostheta, Rsintheta)$.

                Let $E = (a, b)$ be the external point.

                Then the two points $M$ corresponding to a tangeant passing through $E$ are such that vector $vec{CM}$ is orthogonal to $vec{EM}$.



                It immediately gives:
                $$costheta,(x_0+ Rcostheta - a)+ sintheta,(y_0+Rsintheta-b) = 0$$
                And then
                $$costheta,(x_0-a) + sintheta,(y_0-b) + R$$



                One possibility to go further is to rewrite $vec{EC}$ as $(rhocosphi, rhosinphi )$

                And therefore
                $$cos(theta - phi) = -frac{R}{rho}$$






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Let us call $R$ the ray and $C = (x_0, y_0)$ the center of the circle. The points of the circle are the $M =(Rcostheta, Rsintheta)$.

                  Let $E = (a, b)$ be the external point.

                  Then the two points $M$ corresponding to a tangeant passing through $E$ are such that vector $vec{CM}$ is orthogonal to $vec{EM}$.



                  It immediately gives:
                  $$costheta,(x_0+ Rcostheta - a)+ sintheta,(y_0+Rsintheta-b) = 0$$
                  And then
                  $$costheta,(x_0-a) + sintheta,(y_0-b) + R$$



                  One possibility to go further is to rewrite $vec{EC}$ as $(rhocosphi, rhosinphi )$

                  And therefore
                  $$cos(theta - phi) = -frac{R}{rho}$$






                  share|cite|improve this answer











                  $endgroup$



                  Let us call $R$ the ray and $C = (x_0, y_0)$ the center of the circle. The points of the circle are the $M =(Rcostheta, Rsintheta)$.

                  Let $E = (a, b)$ be the external point.

                  Then the two points $M$ corresponding to a tangeant passing through $E$ are such that vector $vec{CM}$ is orthogonal to $vec{EM}$.



                  It immediately gives:
                  $$costheta,(x_0+ Rcostheta - a)+ sintheta,(y_0+Rsintheta-b) = 0$$
                  And then
                  $$costheta,(x_0-a) + sintheta,(y_0-b) + R$$



                  One possibility to go further is to rewrite $vec{EC}$ as $(rhocosphi, rhosinphi )$

                  And therefore
                  $$cos(theta - phi) = -frac{R}{rho}$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 12 at 9:15

























                  answered Jan 12 at 9:04









                  DamienDamien

                  59714




                  59714






























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