Creating a circle equation that has a specific slope at a specific point [closed]












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basically I need to create a circle that will connect with a linear equation at a point so that it is both a continuous and smooth transition, meaning at that point the circle needs to have both the same derivative and y value as the linear equation.
Right now I'm playing around with the equation (ax + b)^2 + (ax + c)^2 = 1
where a b and c are all constants that I'm trying to find out so that when x = 1, y = 4 and the derivative is also 4. I know how to get a line to fit a circle (a tangent) so that its smooth and continuous but I don't know how to create a circle to fit a line, help!










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closed as unclear what you're asking by Brahadeesh, Leucippus, Jean-Claude Arbaut, Rebellos, KReiser Nov 25 '18 at 0:39


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.











  • 3




    $begingroup$
    Not sure this is clear. I think you are saying, you are given a point $P$ on a line $L$ and you want to find a circle tangent to $L$ at $P$. Is this correct? If so, there are infinitely many such circles....just make the perpendicular to $L$ at $P$, take any point on that as the center of your circle. Or did you mean something else?
    $endgroup$
    – lulu
    Nov 24 '18 at 17:18






  • 1




    $begingroup$
    Welcome to math.SE!! Please use MathJax to format your math expressions correctly.
    $endgroup$
    – manooooh
    Nov 24 '18 at 17:20










  • $begingroup$
    Infinitely many circles touch on either side of a given straight line at a given point..
    $endgroup$
    – Narasimham
    Nov 24 '18 at 19:26


















0












$begingroup$


basically I need to create a circle that will connect with a linear equation at a point so that it is both a continuous and smooth transition, meaning at that point the circle needs to have both the same derivative and y value as the linear equation.
Right now I'm playing around with the equation (ax + b)^2 + (ax + c)^2 = 1
where a b and c are all constants that I'm trying to find out so that when x = 1, y = 4 and the derivative is also 4. I know how to get a line to fit a circle (a tangent) so that its smooth and continuous but I don't know how to create a circle to fit a line, help!










share|cite|improve this question









$endgroup$



closed as unclear what you're asking by Brahadeesh, Leucippus, Jean-Claude Arbaut, Rebellos, KReiser Nov 25 '18 at 0:39


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.











  • 3




    $begingroup$
    Not sure this is clear. I think you are saying, you are given a point $P$ on a line $L$ and you want to find a circle tangent to $L$ at $P$. Is this correct? If so, there are infinitely many such circles....just make the perpendicular to $L$ at $P$, take any point on that as the center of your circle. Or did you mean something else?
    $endgroup$
    – lulu
    Nov 24 '18 at 17:18






  • 1




    $begingroup$
    Welcome to math.SE!! Please use MathJax to format your math expressions correctly.
    $endgroup$
    – manooooh
    Nov 24 '18 at 17:20










  • $begingroup$
    Infinitely many circles touch on either side of a given straight line at a given point..
    $endgroup$
    – Narasimham
    Nov 24 '18 at 19:26
















0












0








0





$begingroup$


basically I need to create a circle that will connect with a linear equation at a point so that it is both a continuous and smooth transition, meaning at that point the circle needs to have both the same derivative and y value as the linear equation.
Right now I'm playing around with the equation (ax + b)^2 + (ax + c)^2 = 1
where a b and c are all constants that I'm trying to find out so that when x = 1, y = 4 and the derivative is also 4. I know how to get a line to fit a circle (a tangent) so that its smooth and continuous but I don't know how to create a circle to fit a line, help!










share|cite|improve this question









$endgroup$




basically I need to create a circle that will connect with a linear equation at a point so that it is both a continuous and smooth transition, meaning at that point the circle needs to have both the same derivative and y value as the linear equation.
Right now I'm playing around with the equation (ax + b)^2 + (ax + c)^2 = 1
where a b and c are all constants that I'm trying to find out so that when x = 1, y = 4 and the derivative is also 4. I know how to get a line to fit a circle (a tangent) so that its smooth and continuous but I don't know how to create a circle to fit a line, help!







calculus multivariable-calculus






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asked Nov 24 '18 at 17:15









RADRIRADRI

1




1




closed as unclear what you're asking by Brahadeesh, Leucippus, Jean-Claude Arbaut, Rebellos, KReiser Nov 25 '18 at 0:39


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.






closed as unclear what you're asking by Brahadeesh, Leucippus, Jean-Claude Arbaut, Rebellos, KReiser Nov 25 '18 at 0:39


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.










  • 3




    $begingroup$
    Not sure this is clear. I think you are saying, you are given a point $P$ on a line $L$ and you want to find a circle tangent to $L$ at $P$. Is this correct? If so, there are infinitely many such circles....just make the perpendicular to $L$ at $P$, take any point on that as the center of your circle. Or did you mean something else?
    $endgroup$
    – lulu
    Nov 24 '18 at 17:18






  • 1




    $begingroup$
    Welcome to math.SE!! Please use MathJax to format your math expressions correctly.
    $endgroup$
    – manooooh
    Nov 24 '18 at 17:20










  • $begingroup$
    Infinitely many circles touch on either side of a given straight line at a given point..
    $endgroup$
    – Narasimham
    Nov 24 '18 at 19:26
















  • 3




    $begingroup$
    Not sure this is clear. I think you are saying, you are given a point $P$ on a line $L$ and you want to find a circle tangent to $L$ at $P$. Is this correct? If so, there are infinitely many such circles....just make the perpendicular to $L$ at $P$, take any point on that as the center of your circle. Or did you mean something else?
    $endgroup$
    – lulu
    Nov 24 '18 at 17:18






  • 1




    $begingroup$
    Welcome to math.SE!! Please use MathJax to format your math expressions correctly.
    $endgroup$
    – manooooh
    Nov 24 '18 at 17:20










  • $begingroup$
    Infinitely many circles touch on either side of a given straight line at a given point..
    $endgroup$
    – Narasimham
    Nov 24 '18 at 19:26










3




3




$begingroup$
Not sure this is clear. I think you are saying, you are given a point $P$ on a line $L$ and you want to find a circle tangent to $L$ at $P$. Is this correct? If so, there are infinitely many such circles....just make the perpendicular to $L$ at $P$, take any point on that as the center of your circle. Or did you mean something else?
$endgroup$
– lulu
Nov 24 '18 at 17:18




$begingroup$
Not sure this is clear. I think you are saying, you are given a point $P$ on a line $L$ and you want to find a circle tangent to $L$ at $P$. Is this correct? If so, there are infinitely many such circles....just make the perpendicular to $L$ at $P$, take any point on that as the center of your circle. Or did you mean something else?
$endgroup$
– lulu
Nov 24 '18 at 17:18




1




1




$begingroup$
Welcome to math.SE!! Please use MathJax to format your math expressions correctly.
$endgroup$
– manooooh
Nov 24 '18 at 17:20




$begingroup$
Welcome to math.SE!! Please use MathJax to format your math expressions correctly.
$endgroup$
– manooooh
Nov 24 '18 at 17:20












$begingroup$
Infinitely many circles touch on either side of a given straight line at a given point..
$endgroup$
– Narasimham
Nov 24 '18 at 19:26






$begingroup$
Infinitely many circles touch on either side of a given straight line at a given point..
$endgroup$
– Narasimham
Nov 24 '18 at 19:26












1 Answer
1






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oldest

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2












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HINT



All the tangents circles with center lying on the perpendicular at the tangent line at that point are solutions.



enter image description here






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    HINT



    All the tangents circles with center lying on the perpendicular at the tangent line at that point are solutions.



    enter image description here






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      HINT



      All the tangents circles with center lying on the perpendicular at the tangent line at that point are solutions.



      enter image description here






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        HINT



        All the tangents circles with center lying on the perpendicular at the tangent line at that point are solutions.



        enter image description here






        share|cite|improve this answer









        $endgroup$



        HINT



        All the tangents circles with center lying on the perpendicular at the tangent line at that point are solutions.



        enter image description here







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 24 '18 at 17:20









        gimusigimusi

        1




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