Creating a circle equation that has a specific slope at a specific point [closed]
up vote
0
down vote
favorite
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
asked Nov 24 at 17:15
RADRI
1
closed as unclear what you're asking by Brahadeesh, Leucippus, Jean-Claude Arbaut, Rebellos, KReiser Nov 25 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.
add a comment |
up vote
0
down vote
favorite
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
asked Nov 24 at 17:15
RADRI
1
closed as unclear what you're asking by Brahadeesh, Leucippus, Jean-Claude Arbaut, Rebellos, KReiser Nov 25 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
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?
– lulu
Nov 24 at 17:18
1
Welcome to math.SE!! Please use MathJax to format your math expressions correctly.
– manooooh
Nov 24 at 17:20
Infinitely many circles touch on either side of a given straight line at a given point..
– Narasimham
Nov 24 at 19:26
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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
asked Nov 24 at 17:15
RADRI
1
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
calculus multivariable-calculus
asked Nov 24 at 17:15
RADRI
1
asked Nov 24 at 17:15
RADRI
1
asked Nov 24 at 17:15
RADRI
1
asked Nov 24 at 17:15
RADRI
1
asked Nov 24 at 17:15
RADRI
1
1
closed as unclear what you're asking by Brahadeesh, Leucippus, Jean-Claude Arbaut, Rebellos, KReiser Nov 25 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 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
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?
– lulu
Nov 24 at 17:18
1
Welcome to math.SE!! Please use MathJax to format your math expressions correctly.
– manooooh
Nov 24 at 17:20
Infinitely many circles touch on either side of a given straight line at a given point..
– Narasimham
Nov 24 at 19:26
add a comment |
3
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?
– lulu
Nov 24 at 17:18
1
Welcome to math.SE!! Please use MathJax to format your math expressions correctly.
– manooooh
Nov 24 at 17:20
Infinitely many circles touch on either side of a given straight line at a given point..
– Narasimham
Nov 24 at 19:26
3
3
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?
– lulu
Nov 24 at 17:18
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?
– lulu
Nov 24 at 17:18
1
1
Welcome to math.SE!! Please use MathJax to format your math expressions correctly.
– manooooh
Nov 24 at 17:20
Welcome to math.SE!! Please use MathJax to format your math expressions correctly.
– manooooh
Nov 24 at 17:20
Infinitely many circles touch on either side of a given straight line at a given point..
– Narasimham
Nov 24 at 19:26
Infinitely many circles touch on either side of a given straight line at a given point..
– Narasimham
Nov 24 at 19:26
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
HINT
All the tangents circles with center lying on the perpendicular at the tangent line at that point are solutions.
answered Nov 24 at 17:20
gimusi
90.7k74495
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
HINT
All the tangents circles with center lying on the perpendicular at the tangent line at that point are solutions.
answered Nov 24 at 17:20
gimusi
90.7k74495
add a comment |
up vote
2
down vote
HINT
All the tangents circles with center lying on the perpendicular at the tangent line at that point are solutions.
answered Nov 24 at 17:20
gimusi
90.7k74495
add a comment |
up vote
2
down vote
up vote
2
down vote
HINT
All the tangents circles with center lying on the perpendicular at the tangent line at that point are solutions.
answered Nov 24 at 17:20
gimusi
90.7k74495
HINT
All the tangents circles with center lying on the perpendicular at the tangent line at that point are solutions.
answered Nov 24 at 17:20
gimusi
90.7k74495
answered Nov 24 at 17:20
gimusi
90.7k74495
answered Nov 24 at 17:20
gimusi
90.7k74495
answered Nov 24 at 17:20
gimusi
90.7k74495
90.7k74495
add a comment |
add a comment |
3
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?
– lulu
Nov 24 at 17:18
1
Welcome to math.SE!! Please use MathJax to format your math expressions correctly.
– manooooh
Nov 24 at 17:20
Infinitely many circles touch on either side of a given straight line at a given point..
– Narasimham
Nov 24 at 19:26