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







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  • 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






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share|cite|improve this question













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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





















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    2 Answers
    2






    active

    oldest

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    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









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