Unit circle shifted upwards so it is tangent the graph of $f(x)=x^{2}$












2














How can we find k so that $x^{2}+(y-k)^{2}$ is tangent to the graph of $f(x)=x^2$?Picture






calculus analytic-geometry







share|cite|improve this question




















  • 1




    Barely touches is not a mathematic term. Maybe you mean being tangent to it ?
    – Rebellos
    Nov 28 '18 at 13:02






  • 1




    I think 'touches' suffices?
    – Shubham Johri
    Nov 28 '18 at 13:08
















2














How can we find k so that $x^{2}+(y-k)^{2}$ is tangent to the graph of $f(x)=x^2$?Picture






calculus analytic-geometry







share|cite|improve this question




















  • 1




    Barely touches is not a mathematic term. Maybe you mean being tangent to it ?
    – Rebellos
    Nov 28 '18 at 13:02






  • 1




    I think 'touches' suffices?
    – Shubham Johri
    Nov 28 '18 at 13:08














2












2








2


0





How can we find k so that $x^{2}+(y-k)^{2}$ is tangent to the graph of $f(x)=x^2$?Picture






calculus analytic-geometry







share|cite|improve this question















How can we find k so that $x^{2}+(y-k)^{2}$ is tangent to the graph of $f(x)=x^2$?Picture






calculus analytic-geometry




calculus analytic-geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 '18 at 13:25

























asked Nov 28 '18 at 12:57









Lok

264




264








  • 1




    Barely touches is not a mathematic term. Maybe you mean being tangent to it ?
    – Rebellos
    Nov 28 '18 at 13:02






  • 1




    I think 'touches' suffices?
    – Shubham Johri
    Nov 28 '18 at 13:08














  • 1




    Barely touches is not a mathematic term. Maybe you mean being tangent to it ?
    – Rebellos
    Nov 28 '18 at 13:02






  • 1




    I think 'touches' suffices?
    – Shubham Johri
    Nov 28 '18 at 13:08








1




1




Barely touches is not a mathematic term. Maybe you mean being tangent to it ?
– Rebellos
Nov 28 '18 at 13:02




Barely touches is not a mathematic term. Maybe you mean being tangent to it ?
– Rebellos
Nov 28 '18 at 13:02




1




1




I think 'touches' suffices?
– Shubham Johri
Nov 28 '18 at 13:08




I think 'touches' suffices?
– Shubham Johri
Nov 28 '18 at 13:08










2 Answers
2






active

oldest

votes


















2














Substantial hint:



Suppose you knew $k$. Then at the (right hand) point where the graphs touch -- call it $(a,b)$, you'd know three things:




  1. $b = a^2$

  2. $a^2 + (b-k)^2 = 1$

  3. $frac{b-k}{a} = frac{-1}{2a}$


The last follows by drawing a line from $(a, b)$ to the circle center at $(0, k)$. That line has slope $frac{b-k}{a}$. It's also perpendicular to the tangent line to the parabola at $(a, b)$, which has slope $2a$. And the product of the slopes of perpendicular lines is always $-1$.



Now you just have to solve those three simultaneous equations.






share|cite|improve this answer





















  • I got $k=frac{7}{8}\$, but when I plot it it doesn't seem right.
    – Lok
    Nov 28 '18 at 14:41








  • 1




    Are you sure that you've made calculations right? I got $k=frac74$ in my mind (unfortunately I do not have a paper and pen right now)
    – Mikalai Parshutsich
    Nov 28 '18 at 15:49






  • 1




    When you got $k = 7/8$, what values did you get for $a$ and $b$? Do the resulting values satisfy all three equations? If not, then you made a mistake.
    – John Hughes
    Nov 28 '18 at 15:54






  • 1




    You could even add your method for arriving at $7/8$ to your question (click on "edit" just below the question itself) so that we might show where you went off the rails.
    – John Hughes
    Nov 28 '18 at 16:44










  • Made a mistake. Sorry for bothering. The answer is $frac{5}{4}$.
    – Lok
    Nov 28 '18 at 22:38



















1














The points of intersection of the parabola and the circle are the solutions of the pair of equations $$y=x^2 text{ and } x^{2}+(y-k)^{2}=1 $$ or, equivalently, $$y=x^2 text{ and } y+(y-k)^{2}=1 $$
The circle is tangent to the parabola if the quadratic equation for $y$ has exactly one solution.






share|cite|improve this answer





















    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3017108%2funit-circle-shifted-upwards-so-it-is-tangent-the-graph-of-fx-x2%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    Substantial hint:



    Suppose you knew $k$. Then at the (right hand) point where the graphs touch -- call it $(a,b)$, you'd know three things:




    1. $b = a^2$

    2. $a^2 + (b-k)^2 = 1$

    3. $frac{b-k}{a} = frac{-1}{2a}$


    The last follows by drawing a line from $(a, b)$ to the circle center at $(0, k)$. That line has slope $frac{b-k}{a}$. It's also perpendicular to the tangent line to the parabola at $(a, b)$, which has slope $2a$. And the product of the slopes of perpendicular lines is always $-1$.



    Now you just have to solve those three simultaneous equations.






    share|cite|improve this answer





















    • I got $k=frac{7}{8}\$, but when I plot it it doesn't seem right.
      – Lok
      Nov 28 '18 at 14:41








    • 1




      Are you sure that you've made calculations right? I got $k=frac74$ in my mind (unfortunately I do not have a paper and pen right now)
      – Mikalai Parshutsich
      Nov 28 '18 at 15:49






    • 1




      When you got $k = 7/8$, what values did you get for $a$ and $b$? Do the resulting values satisfy all three equations? If not, then you made a mistake.
      – John Hughes
      Nov 28 '18 at 15:54






    • 1




      You could even add your method for arriving at $7/8$ to your question (click on "edit" just below the question itself) so that we might show where you went off the rails.
      – John Hughes
      Nov 28 '18 at 16:44










    • Made a mistake. Sorry for bothering. The answer is $frac{5}{4}$.
      – Lok
      Nov 28 '18 at 22:38
















    2














    Substantial hint:



    Suppose you knew $k$. Then at the (right hand) point where the graphs touch -- call it $(a,b)$, you'd know three things:




    1. $b = a^2$

    2. $a^2 + (b-k)^2 = 1$

    3. $frac{b-k}{a} = frac{-1}{2a}$


    The last follows by drawing a line from $(a, b)$ to the circle center at $(0, k)$. That line has slope $frac{b-k}{a}$. It's also perpendicular to the tangent line to the parabola at $(a, b)$, which has slope $2a$. And the product of the slopes of perpendicular lines is always $-1$.



    Now you just have to solve those three simultaneous equations.






    share|cite|improve this answer





















    • I got $k=frac{7}{8}\$, but when I plot it it doesn't seem right.
      – Lok
      Nov 28 '18 at 14:41








    • 1




      Are you sure that you've made calculations right? I got $k=frac74$ in my mind (unfortunately I do not have a paper and pen right now)
      – Mikalai Parshutsich
      Nov 28 '18 at 15:49






    • 1




      When you got $k = 7/8$, what values did you get for $a$ and $b$? Do the resulting values satisfy all three equations? If not, then you made a mistake.
      – John Hughes
      Nov 28 '18 at 15:54






    • 1




      You could even add your method for arriving at $7/8$ to your question (click on "edit" just below the question itself) so that we might show where you went off the rails.
      – John Hughes
      Nov 28 '18 at 16:44










    • Made a mistake. Sorry for bothering. The answer is $frac{5}{4}$.
      – Lok
      Nov 28 '18 at 22:38














    2












    2








    2






    Substantial hint:



    Suppose you knew $k$. Then at the (right hand) point where the graphs touch -- call it $(a,b)$, you'd know three things:




    1. $b = a^2$

    2. $a^2 + (b-k)^2 = 1$

    3. $frac{b-k}{a} = frac{-1}{2a}$


    The last follows by drawing a line from $(a, b)$ to the circle center at $(0, k)$. That line has slope $frac{b-k}{a}$. It's also perpendicular to the tangent line to the parabola at $(a, b)$, which has slope $2a$. And the product of the slopes of perpendicular lines is always $-1$.



    Now you just have to solve those three simultaneous equations.






    share|cite|improve this answer












    Substantial hint:



    Suppose you knew $k$. Then at the (right hand) point where the graphs touch -- call it $(a,b)$, you'd know three things:




    1. $b = a^2$

    2. $a^2 + (b-k)^2 = 1$

    3. $frac{b-k}{a} = frac{-1}{2a}$


    The last follows by drawing a line from $(a, b)$ to the circle center at $(0, k)$. That line has slope $frac{b-k}{a}$. It's also perpendicular to the tangent line to the parabola at $(a, b)$, which has slope $2a$. And the product of the slopes of perpendicular lines is always $-1$.



    Now you just have to solve those three simultaneous equations.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 28 '18 at 13:26









    John Hughes

    62.5k24090




    62.5k24090












    • I got $k=frac{7}{8}\$, but when I plot it it doesn't seem right.
      – Lok
      Nov 28 '18 at 14:41








    • 1




      Are you sure that you've made calculations right? I got $k=frac74$ in my mind (unfortunately I do not have a paper and pen right now)
      – Mikalai Parshutsich
      Nov 28 '18 at 15:49






    • 1




      When you got $k = 7/8$, what values did you get for $a$ and $b$? Do the resulting values satisfy all three equations? If not, then you made a mistake.
      – John Hughes
      Nov 28 '18 at 15:54






    • 1




      You could even add your method for arriving at $7/8$ to your question (click on "edit" just below the question itself) so that we might show where you went off the rails.
      – John Hughes
      Nov 28 '18 at 16:44










    • Made a mistake. Sorry for bothering. The answer is $frac{5}{4}$.
      – Lok
      Nov 28 '18 at 22:38


















    • I got $k=frac{7}{8}\$, but when I plot it it doesn't seem right.
      – Lok
      Nov 28 '18 at 14:41








    • 1




      Are you sure that you've made calculations right? I got $k=frac74$ in my mind (unfortunately I do not have a paper and pen right now)
      – Mikalai Parshutsich
      Nov 28 '18 at 15:49






    • 1




      When you got $k = 7/8$, what values did you get for $a$ and $b$? Do the resulting values satisfy all three equations? If not, then you made a mistake.
      – John Hughes
      Nov 28 '18 at 15:54






    • 1




      You could even add your method for arriving at $7/8$ to your question (click on "edit" just below the question itself) so that we might show where you went off the rails.
      – John Hughes
      Nov 28 '18 at 16:44










    • Made a mistake. Sorry for bothering. The answer is $frac{5}{4}$.
      – Lok
      Nov 28 '18 at 22:38
















    I got $k=frac{7}{8}\$, but when I plot it it doesn't seem right.
    – Lok
    Nov 28 '18 at 14:41






    I got $k=frac{7}{8}\$, but when I plot it it doesn't seem right.
    – Lok
    Nov 28 '18 at 14:41






    1




    1




    Are you sure that you've made calculations right? I got $k=frac74$ in my mind (unfortunately I do not have a paper and pen right now)
    – Mikalai Parshutsich
    Nov 28 '18 at 15:49




    Are you sure that you've made calculations right? I got $k=frac74$ in my mind (unfortunately I do not have a paper and pen right now)
    – Mikalai Parshutsich
    Nov 28 '18 at 15:49




    1




    1




    When you got $k = 7/8$, what values did you get for $a$ and $b$? Do the resulting values satisfy all three equations? If not, then you made a mistake.
    – John Hughes
    Nov 28 '18 at 15:54




    When you got $k = 7/8$, what values did you get for $a$ and $b$? Do the resulting values satisfy all three equations? If not, then you made a mistake.
    – John Hughes
    Nov 28 '18 at 15:54




    1




    1




    You could even add your method for arriving at $7/8$ to your question (click on "edit" just below the question itself) so that we might show where you went off the rails.
    – John Hughes
    Nov 28 '18 at 16:44




    You could even add your method for arriving at $7/8$ to your question (click on "edit" just below the question itself) so that we might show where you went off the rails.
    – John Hughes
    Nov 28 '18 at 16:44












    Made a mistake. Sorry for bothering. The answer is $frac{5}{4}$.
    – Lok
    Nov 28 '18 at 22:38




    Made a mistake. Sorry for bothering. The answer is $frac{5}{4}$.
    – Lok
    Nov 28 '18 at 22:38











    1














    The points of intersection of the parabola and the circle are the solutions of the pair of equations $$y=x^2 text{ and } x^{2}+(y-k)^{2}=1 $$ or, equivalently, $$y=x^2 text{ and } y+(y-k)^{2}=1 $$
    The circle is tangent to the parabola if the quadratic equation for $y$ has exactly one solution.






    share|cite|improve this answer


























      1














      The points of intersection of the parabola and the circle are the solutions of the pair of equations $$y=x^2 text{ and } x^{2}+(y-k)^{2}=1 $$ or, equivalently, $$y=x^2 text{ and } y+(y-k)^{2}=1 $$
      The circle is tangent to the parabola if the quadratic equation for $y$ has exactly one solution.






      share|cite|improve this answer
























        1












        1








        1






        The points of intersection of the parabola and the circle are the solutions of the pair of equations $$y=x^2 text{ and } x^{2}+(y-k)^{2}=1 $$ or, equivalently, $$y=x^2 text{ and } y+(y-k)^{2}=1 $$
        The circle is tangent to the parabola if the quadratic equation for $y$ has exactly one solution.






        share|cite|improve this answer












        The points of intersection of the parabola and the circle are the solutions of the pair of equations $$y=x^2 text{ and } x^{2}+(y-k)^{2}=1 $$ or, equivalently, $$y=x^2 text{ and } y+(y-k)^{2}=1 $$
        The circle is tangent to the parabola if the quadratic equation for $y$ has exactly one solution.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 '18 at 23:15









        random

        51626




        51626






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3017108%2funit-circle-shifted-upwards-so-it-is-tangent-the-graph-of-fx-x2%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Ellipse (mathématiques)

            Image
            Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
            Read more

            Quarter-circle Tiles

            Image
            up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
            Read more

            Mont Emei

            Image
            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
            Read more