A circle of radius $r$ is dropped into the parabola $y=x^{2}$. Find the largest $r$ so the circle will touch...












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If $r$ is too large, the circle will not fall to the bottom, if $r$ is sufficiently small, the circle will touch the parabola at its vertex $(0, 0)$. Find the largest value of $r$ s.t. the circle will touch the vertex of the parabola.






calculus geometry functions







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  • 2




    $begingroup$
    That is equivalent to finding the osculating circle at the vertex of the parabola, i.e. to computing a curvature. What have you attempted?
    $endgroup$
    – Jack D'Aurizio
    Aug 19 '15 at 9:29










  • $begingroup$
    I thought that I would need to somehow use the equation of the circle to find the circle that intersects the parabola at the origin while pointing up the $y$ direction. That is $(x -0)^{2} - (y-h)^{2} = r^{2}$. I know that $x$ must be at $0$ but I don't know that value of $y$ nor the radius.
    $endgroup$
    – user3063381
    Aug 19 '15 at 9:41
















2












$begingroup$


If $r$ is too large, the circle will not fall to the bottom, if $r$ is sufficiently small, the circle will touch the parabola at its vertex $(0, 0)$. Find the largest value of $r$ s.t. the circle will touch the vertex of the parabola.






calculus geometry functions







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    That is equivalent to finding the osculating circle at the vertex of the parabola, i.e. to computing a curvature. What have you attempted?
    $endgroup$
    – Jack D'Aurizio
    Aug 19 '15 at 9:29










  • $begingroup$
    I thought that I would need to somehow use the equation of the circle to find the circle that intersects the parabola at the origin while pointing up the $y$ direction. That is $(x -0)^{2} - (y-h)^{2} = r^{2}$. I know that $x$ must be at $0$ but I don't know that value of $y$ nor the radius.
    $endgroup$
    – user3063381
    Aug 19 '15 at 9:41














2












2








2


1



$begingroup$


If $r$ is too large, the circle will not fall to the bottom, if $r$ is sufficiently small, the circle will touch the parabola at its vertex $(0, 0)$. Find the largest value of $r$ s.t. the circle will touch the vertex of the parabola.






calculus geometry functions







share|cite|improve this question









$endgroup$




If $r$ is too large, the circle will not fall to the bottom, if $r$ is sufficiently small, the circle will touch the parabola at its vertex $(0, 0)$. Find the largest value of $r$ s.t. the circle will touch the vertex of the parabola.






calculus geometry functions




calculus geometry functions






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asked Aug 19 '15 at 9:27









user3063381user3063381

446




446








  • 2




    $begingroup$
    That is equivalent to finding the osculating circle at the vertex of the parabola, i.e. to computing a curvature. What have you attempted?
    $endgroup$
    – Jack D'Aurizio
    Aug 19 '15 at 9:29










  • $begingroup$
    I thought that I would need to somehow use the equation of the circle to find the circle that intersects the parabola at the origin while pointing up the $y$ direction. That is $(x -0)^{2} - (y-h)^{2} = r^{2}$. I know that $x$ must be at $0$ but I don't know that value of $y$ nor the radius.
    $endgroup$
    – user3063381
    Aug 19 '15 at 9:41














  • 2




    $begingroup$
    That is equivalent to finding the osculating circle at the vertex of the parabola, i.e. to computing a curvature. What have you attempted?
    $endgroup$
    – Jack D'Aurizio
    Aug 19 '15 at 9:29










  • $begingroup$
    I thought that I would need to somehow use the equation of the circle to find the circle that intersects the parabola at the origin while pointing up the $y$ direction. That is $(x -0)^{2} - (y-h)^{2} = r^{2}$. I know that $x$ must be at $0$ but I don't know that value of $y$ nor the radius.
    $endgroup$
    – user3063381
    Aug 19 '15 at 9:41








2




2




$begingroup$
That is equivalent to finding the osculating circle at the vertex of the parabola, i.e. to computing a curvature. What have you attempted?
$endgroup$
– Jack D'Aurizio
Aug 19 '15 at 9:29




$begingroup$
That is equivalent to finding the osculating circle at the vertex of the parabola, i.e. to computing a curvature. What have you attempted?
$endgroup$
– Jack D'Aurizio
Aug 19 '15 at 9:29












$begingroup$
I thought that I would need to somehow use the equation of the circle to find the circle that intersects the parabola at the origin while pointing up the $y$ direction. That is $(x -0)^{2} - (y-h)^{2} = r^{2}$. I know that $x$ must be at $0$ but I don't know that value of $y$ nor the radius.
$endgroup$
– user3063381
Aug 19 '15 at 9:41




$begingroup$
I thought that I would need to somehow use the equation of the circle to find the circle that intersects the parabola at the origin while pointing up the $y$ direction. That is $(x -0)^{2} - (y-h)^{2} = r^{2}$. I know that $x$ must be at $0$ but I don't know that value of $y$ nor the radius.
$endgroup$
– user3063381
Aug 19 '15 at 9:41










4 Answers
4






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$begingroup$

Let the center of the circle $(0, r)$ & radius $r$ hence the equation of the circle $$(x-0)^2+(y-r)^2=r^2$$ $$x^2+(y-r)^2=r^2$$ Now, solving the equations of parabola $y=x^2$ & equation of the circle we get $$x^2+(x^2-r)^2=r^2$$ $$x^2+x^4+r^2-2rx^2=r^2$$ $$x^4-(2r-1)x^2=0$$ $$x^2(x^2+1-2r)=0$$
$$x^2=0 vee x^2+1-2r=0$$ $$x=0 vee x^2=2r-1$$
$$x=0 vee x^=pmsqrt{2r-1}$$ Corresponding values of $y$ are calculated as follows

$$(x=0, y=0), (x=sqrt{2r-1}, y=2r-1), (x=-sqrt{2r-1}, y=2r-1)$$



Thus, we find the three points of intersection of the circle with the parabola $(0, 0)$, $(sqrt{2r-1}, 2r-1)$ & $(-sqrt{2r-1}, 2r-1)$



But the circle is touching the parabola at the vertex $(0, 0)$ only hence, it is possible when other two points $(sqrt{2r-1}, 2r-1)$ & $(-sqrt{2r-1}, 2r-1)$ are coinciding with the vertex $(0, 0)$ hence, we get $$pm sqrt{2r-1}=0iff r=frac{1}{2}$$
& $$2r-1=0iff r=frac{1}{2}$$
$$bbox[5px, border:2px solid #C0A000]{color{red}{text{Maximum radius of circle, }r_{text{max}}=color{blue}{frac{1}{2}}}}$$






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    Following my comment, you just have to compute for which $r$s
    $$ f(x)=sqrt{r^2-x^2} = r-x^2 = g(x) $$
    just at $x=0$. Since:
    $$ (r^2-x^2)-(r-x^2)^2 = x^2(2r-1-x^2) $$
    the critical radius is obviously $r=frac{1}{2}$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Could you explain the first equation?
      $endgroup$
      – user3063381
      Aug 19 '15 at 9:44










    • $begingroup$
      @user3063381: $f$ and $g$ represent the semicircle having center in the origin through $(0,r)$ and the parabola $y=x^2$ reflected with respect to $(0,r/2)$. So, instead of dropping a circle inside a parabola, I dropped a parabola with fixed $a$ on a circle with radius $r$.
      $endgroup$
      – Jack D'Aurizio
      Aug 19 '15 at 9:47





















    1












    $begingroup$

    Maximum radius is also the curvature at x=0.



    Curvature of a plane curve is d/dx(dy/dx)/(1 + (dy/dx)^2)^ 3/2



    For Parabola y = x^2,



    The value of curvature is 2. Therefore R = 1/2






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Idea is similar with the accepted answer, yet it is simpler.



      The equation of the circle with center $(0,r)$ is: $(x-0)^2+(y-r)^2=r^2$.



      The equation of the parabola is: $y=x^2$.



      The two graphs must have a single point of contact at: $(0,0)$.



      Substitute $y=x^2$ into the equation of circle:
      $$y+(y-r)^2=r^2 Rightarrow y^2+(1-2r)y=0 Rightarrow y_1=0; y_2=2r-1.$$
      We find:
      $$x^2=0 Rightarrow x_1=0;\
      x^2=2r-1 le 0 Rightarrow rle frac12.$$






      share|cite|improve this answer









      $endgroup$














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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        Let the center of the circle $(0, r)$ & radius $r$ hence the equation of the circle $$(x-0)^2+(y-r)^2=r^2$$ $$x^2+(y-r)^2=r^2$$ Now, solving the equations of parabola $y=x^2$ & equation of the circle we get $$x^2+(x^2-r)^2=r^2$$ $$x^2+x^4+r^2-2rx^2=r^2$$ $$x^4-(2r-1)x^2=0$$ $$x^2(x^2+1-2r)=0$$
        $$x^2=0 vee x^2+1-2r=0$$ $$x=0 vee x^2=2r-1$$
        $$x=0 vee x^=pmsqrt{2r-1}$$ Corresponding values of $y$ are calculated as follows

        $$(x=0, y=0), (x=sqrt{2r-1}, y=2r-1), (x=-sqrt{2r-1}, y=2r-1)$$



        Thus, we find the three points of intersection of the circle with the parabola $(0, 0)$, $(sqrt{2r-1}, 2r-1)$ & $(-sqrt{2r-1}, 2r-1)$



        But the circle is touching the parabola at the vertex $(0, 0)$ only hence, it is possible when other two points $(sqrt{2r-1}, 2r-1)$ & $(-sqrt{2r-1}, 2r-1)$ are coinciding with the vertex $(0, 0)$ hence, we get $$pm sqrt{2r-1}=0iff r=frac{1}{2}$$
        & $$2r-1=0iff r=frac{1}{2}$$
        $$bbox[5px, border:2px solid #C0A000]{color{red}{text{Maximum radius of circle, }r_{text{max}}=color{blue}{frac{1}{2}}}}$$






        share|cite|improve this answer











        $endgroup$


















          2












          $begingroup$

          Let the center of the circle $(0, r)$ & radius $r$ hence the equation of the circle $$(x-0)^2+(y-r)^2=r^2$$ $$x^2+(y-r)^2=r^2$$ Now, solving the equations of parabola $y=x^2$ & equation of the circle we get $$x^2+(x^2-r)^2=r^2$$ $$x^2+x^4+r^2-2rx^2=r^2$$ $$x^4-(2r-1)x^2=0$$ $$x^2(x^2+1-2r)=0$$
          $$x^2=0 vee x^2+1-2r=0$$ $$x=0 vee x^2=2r-1$$
          $$x=0 vee x^=pmsqrt{2r-1}$$ Corresponding values of $y$ are calculated as follows

          $$(x=0, y=0), (x=sqrt{2r-1}, y=2r-1), (x=-sqrt{2r-1}, y=2r-1)$$



          Thus, we find the three points of intersection of the circle with the parabola $(0, 0)$, $(sqrt{2r-1}, 2r-1)$ & $(-sqrt{2r-1}, 2r-1)$



          But the circle is touching the parabola at the vertex $(0, 0)$ only hence, it is possible when other two points $(sqrt{2r-1}, 2r-1)$ & $(-sqrt{2r-1}, 2r-1)$ are coinciding with the vertex $(0, 0)$ hence, we get $$pm sqrt{2r-1}=0iff r=frac{1}{2}$$
          & $$2r-1=0iff r=frac{1}{2}$$
          $$bbox[5px, border:2px solid #C0A000]{color{red}{text{Maximum radius of circle, }r_{text{max}}=color{blue}{frac{1}{2}}}}$$






          share|cite|improve this answer











          $endgroup$
















            2












            2








            2





            $begingroup$

            Let the center of the circle $(0, r)$ & radius $r$ hence the equation of the circle $$(x-0)^2+(y-r)^2=r^2$$ $$x^2+(y-r)^2=r^2$$ Now, solving the equations of parabola $y=x^2$ & equation of the circle we get $$x^2+(x^2-r)^2=r^2$$ $$x^2+x^4+r^2-2rx^2=r^2$$ $$x^4-(2r-1)x^2=0$$ $$x^2(x^2+1-2r)=0$$
            $$x^2=0 vee x^2+1-2r=0$$ $$x=0 vee x^2=2r-1$$
            $$x=0 vee x^=pmsqrt{2r-1}$$ Corresponding values of $y$ are calculated as follows

            $$(x=0, y=0), (x=sqrt{2r-1}, y=2r-1), (x=-sqrt{2r-1}, y=2r-1)$$



            Thus, we find the three points of intersection of the circle with the parabola $(0, 0)$, $(sqrt{2r-1}, 2r-1)$ & $(-sqrt{2r-1}, 2r-1)$



            But the circle is touching the parabola at the vertex $(0, 0)$ only hence, it is possible when other two points $(sqrt{2r-1}, 2r-1)$ & $(-sqrt{2r-1}, 2r-1)$ are coinciding with the vertex $(0, 0)$ hence, we get $$pm sqrt{2r-1}=0iff r=frac{1}{2}$$
            & $$2r-1=0iff r=frac{1}{2}$$
            $$bbox[5px, border:2px solid #C0A000]{color{red}{text{Maximum radius of circle, }r_{text{max}}=color{blue}{frac{1}{2}}}}$$






            share|cite|improve this answer











            $endgroup$



            Let the center of the circle $(0, r)$ & radius $r$ hence the equation of the circle $$(x-0)^2+(y-r)^2=r^2$$ $$x^2+(y-r)^2=r^2$$ Now, solving the equations of parabola $y=x^2$ & equation of the circle we get $$x^2+(x^2-r)^2=r^2$$ $$x^2+x^4+r^2-2rx^2=r^2$$ $$x^4-(2r-1)x^2=0$$ $$x^2(x^2+1-2r)=0$$
            $$x^2=0 vee x^2+1-2r=0$$ $$x=0 vee x^2=2r-1$$
            $$x=0 vee x^=pmsqrt{2r-1}$$ Corresponding values of $y$ are calculated as follows

            $$(x=0, y=0), (x=sqrt{2r-1}, y=2r-1), (x=-sqrt{2r-1}, y=2r-1)$$



            Thus, we find the three points of intersection of the circle with the parabola $(0, 0)$, $(sqrt{2r-1}, 2r-1)$ & $(-sqrt{2r-1}, 2r-1)$



            But the circle is touching the parabola at the vertex $(0, 0)$ only hence, it is possible when other two points $(sqrt{2r-1}, 2r-1)$ & $(-sqrt{2r-1}, 2r-1)$ are coinciding with the vertex $(0, 0)$ hence, we get $$pm sqrt{2r-1}=0iff r=frac{1}{2}$$
            & $$2r-1=0iff r=frac{1}{2}$$
            $$bbox[5px, border:2px solid #C0A000]{color{red}{text{Maximum radius of circle, }r_{text{max}}=color{blue}{frac{1}{2}}}}$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Aug 19 '15 at 10:06

























            answered Aug 19 '15 at 9:53









            Harish Chandra RajpootHarish Chandra Rajpoot

            29.7k103772




            29.7k103772























                2












                $begingroup$

                Following my comment, you just have to compute for which $r$s
                $$ f(x)=sqrt{r^2-x^2} = r-x^2 = g(x) $$
                just at $x=0$. Since:
                $$ (r^2-x^2)-(r-x^2)^2 = x^2(2r-1-x^2) $$
                the critical radius is obviously $r=frac{1}{2}$.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  Could you explain the first equation?
                  $endgroup$
                  – user3063381
                  Aug 19 '15 at 9:44










                • $begingroup$
                  @user3063381: $f$ and $g$ represent the semicircle having center in the origin through $(0,r)$ and the parabola $y=x^2$ reflected with respect to $(0,r/2)$. So, instead of dropping a circle inside a parabola, I dropped a parabola with fixed $a$ on a circle with radius $r$.
                  $endgroup$
                  – Jack D'Aurizio
                  Aug 19 '15 at 9:47


















                2












                $begingroup$

                Following my comment, you just have to compute for which $r$s
                $$ f(x)=sqrt{r^2-x^2} = r-x^2 = g(x) $$
                just at $x=0$. Since:
                $$ (r^2-x^2)-(r-x^2)^2 = x^2(2r-1-x^2) $$
                the critical radius is obviously $r=frac{1}{2}$.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  Could you explain the first equation?
                  $endgroup$
                  – user3063381
                  Aug 19 '15 at 9:44










                • $begingroup$
                  @user3063381: $f$ and $g$ represent the semicircle having center in the origin through $(0,r)$ and the parabola $y=x^2$ reflected with respect to $(0,r/2)$. So, instead of dropping a circle inside a parabola, I dropped a parabola with fixed $a$ on a circle with radius $r$.
                  $endgroup$
                  – Jack D'Aurizio
                  Aug 19 '15 at 9:47
















                2












                2








                2





                $begingroup$

                Following my comment, you just have to compute for which $r$s
                $$ f(x)=sqrt{r^2-x^2} = r-x^2 = g(x) $$
                just at $x=0$. Since:
                $$ (r^2-x^2)-(r-x^2)^2 = x^2(2r-1-x^2) $$
                the critical radius is obviously $r=frac{1}{2}$.






                share|cite|improve this answer









                $endgroup$



                Following my comment, you just have to compute for which $r$s
                $$ f(x)=sqrt{r^2-x^2} = r-x^2 = g(x) $$
                just at $x=0$. Since:
                $$ (r^2-x^2)-(r-x^2)^2 = x^2(2r-1-x^2) $$
                the critical radius is obviously $r=frac{1}{2}$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 19 '15 at 9:35









                Jack D'AurizioJack D'Aurizio

                292k33285674




                292k33285674












                • $begingroup$
                  Could you explain the first equation?
                  $endgroup$
                  – user3063381
                  Aug 19 '15 at 9:44










                • $begingroup$
                  @user3063381: $f$ and $g$ represent the semicircle having center in the origin through $(0,r)$ and the parabola $y=x^2$ reflected with respect to $(0,r/2)$. So, instead of dropping a circle inside a parabola, I dropped a parabola with fixed $a$ on a circle with radius $r$.
                  $endgroup$
                  – Jack D'Aurizio
                  Aug 19 '15 at 9:47




















                • $begingroup$
                  Could you explain the first equation?
                  $endgroup$
                  – user3063381
                  Aug 19 '15 at 9:44










                • $begingroup$
                  @user3063381: $f$ and $g$ represent the semicircle having center in the origin through $(0,r)$ and the parabola $y=x^2$ reflected with respect to $(0,r/2)$. So, instead of dropping a circle inside a parabola, I dropped a parabola with fixed $a$ on a circle with radius $r$.
                  $endgroup$
                  – Jack D'Aurizio
                  Aug 19 '15 at 9:47


















                $begingroup$
                Could you explain the first equation?
                $endgroup$
                – user3063381
                Aug 19 '15 at 9:44




                $begingroup$
                Could you explain the first equation?
                $endgroup$
                – user3063381
                Aug 19 '15 at 9:44












                $begingroup$
                @user3063381: $f$ and $g$ represent the semicircle having center in the origin through $(0,r)$ and the parabola $y=x^2$ reflected with respect to $(0,r/2)$. So, instead of dropping a circle inside a parabola, I dropped a parabola with fixed $a$ on a circle with radius $r$.
                $endgroup$
                – Jack D'Aurizio
                Aug 19 '15 at 9:47






                $begingroup$
                @user3063381: $f$ and $g$ represent the semicircle having center in the origin through $(0,r)$ and the parabola $y=x^2$ reflected with respect to $(0,r/2)$. So, instead of dropping a circle inside a parabola, I dropped a parabola with fixed $a$ on a circle with radius $r$.
                $endgroup$
                – Jack D'Aurizio
                Aug 19 '15 at 9:47













                1












                $begingroup$

                Maximum radius is also the curvature at x=0.



                Curvature of a plane curve is d/dx(dy/dx)/(1 + (dy/dx)^2)^ 3/2



                For Parabola y = x^2,



                The value of curvature is 2. Therefore R = 1/2






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Maximum radius is also the curvature at x=0.



                  Curvature of a plane curve is d/dx(dy/dx)/(1 + (dy/dx)^2)^ 3/2



                  For Parabola y = x^2,



                  The value of curvature is 2. Therefore R = 1/2






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Maximum radius is also the curvature at x=0.



                    Curvature of a plane curve is d/dx(dy/dx)/(1 + (dy/dx)^2)^ 3/2



                    For Parabola y = x^2,



                    The value of curvature is 2. Therefore R = 1/2






                    share|cite|improve this answer









                    $endgroup$



                    Maximum radius is also the curvature at x=0.



                    Curvature of a plane curve is d/dx(dy/dx)/(1 + (dy/dx)^2)^ 3/2



                    For Parabola y = x^2,



                    The value of curvature is 2. Therefore R = 1/2







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 24 '18 at 23:21









                    Venkatesan GopalakrishnanVenkatesan Gopalakrishnan

                    111




                    111























                        0












                        $begingroup$

                        Idea is similar with the accepted answer, yet it is simpler.



                        The equation of the circle with center $(0,r)$ is: $(x-0)^2+(y-r)^2=r^2$.



                        The equation of the parabola is: $y=x^2$.



                        The two graphs must have a single point of contact at: $(0,0)$.



                        Substitute $y=x^2$ into the equation of circle:
                        $$y+(y-r)^2=r^2 Rightarrow y^2+(1-2r)y=0 Rightarrow y_1=0; y_2=2r-1.$$
                        We find:
                        $$x^2=0 Rightarrow x_1=0;\
                        x^2=2r-1 le 0 Rightarrow rle frac12.$$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Idea is similar with the accepted answer, yet it is simpler.



                          The equation of the circle with center $(0,r)$ is: $(x-0)^2+(y-r)^2=r^2$.



                          The equation of the parabola is: $y=x^2$.



                          The two graphs must have a single point of contact at: $(0,0)$.



                          Substitute $y=x^2$ into the equation of circle:
                          $$y+(y-r)^2=r^2 Rightarrow y^2+(1-2r)y=0 Rightarrow y_1=0; y_2=2r-1.$$
                          We find:
                          $$x^2=0 Rightarrow x_1=0;\
                          x^2=2r-1 le 0 Rightarrow rle frac12.$$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Idea is similar with the accepted answer, yet it is simpler.



                            The equation of the circle with center $(0,r)$ is: $(x-0)^2+(y-r)^2=r^2$.



                            The equation of the parabola is: $y=x^2$.



                            The two graphs must have a single point of contact at: $(0,0)$.



                            Substitute $y=x^2$ into the equation of circle:
                            $$y+(y-r)^2=r^2 Rightarrow y^2+(1-2r)y=0 Rightarrow y_1=0; y_2=2r-1.$$
                            We find:
                            $$x^2=0 Rightarrow x_1=0;\
                            x^2=2r-1 le 0 Rightarrow rle frac12.$$






                            share|cite|improve this answer









                            $endgroup$



                            Idea is similar with the accepted answer, yet it is simpler.



                            The equation of the circle with center $(0,r)$ is: $(x-0)^2+(y-r)^2=r^2$.



                            The equation of the parabola is: $y=x^2$.



                            The two graphs must have a single point of contact at: $(0,0)$.



                            Substitute $y=x^2$ into the equation of circle:
                            $$y+(y-r)^2=r^2 Rightarrow y^2+(1-2r)y=0 Rightarrow y_1=0; y_2=2r-1.$$
                            We find:
                            $$x^2=0 Rightarrow x_1=0;\
                            x^2=2r-1 le 0 Rightarrow rle frac12.$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 25 '18 at 5:56









                            farruhotafarruhota

                            22.4k2942




                            22.4k2942






























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