How to find a tangent to a circle from an external point using calculus?
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So I know how to find the tangent from an external point using algebra but that involves many equations making the entire process tedious. Anyways I have a calculus exam coming up and I think I should be using calculus to solve such problems .
calculus geometry
asked Jan 12 at 6:34
Ishaan ParikhIshaan Parikh
256
$endgroup$
add a comment |
$begingroup$
So I know how to find the tangent from an external point using algebra but that involves many equations making the entire process tedious. Anyways I have a calculus exam coming up and I think I should be using calculus to solve such problems .
calculus geometry
asked Jan 12 at 6:34
Ishaan ParikhIshaan Parikh
256
$endgroup$
$begingroup$
How many equations do you consider “many?” If you can write down the equation of the circle, you can also write down the equation of the polar of a point. The points of tangency are the solutions to this system of two equations. That doesn’t sound particularly tedious to me.
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– amd
Jan 12 at 9:32
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Why do we need calculus to begin with? Analytic geometry works perfectly well.
$endgroup$
– Oscar Lanzi
Jan 12 at 11:11
add a comment |
$begingroup$
So I know how to find the tangent from an external point using algebra but that involves many equations making the entire process tedious. Anyways I have a calculus exam coming up and I think I should be using calculus to solve such problems .
calculus geometry
asked Jan 12 at 6:34
Ishaan ParikhIshaan Parikh
256
$endgroup$
So I know how to find the tangent from an external point using algebra but that involves many equations making the entire process tedious. Anyways I have a calculus exam coming up and I think I should be using calculus to solve such problems .
calculus geometry
calculus geometry
asked Jan 12 at 6:34
Ishaan ParikhIshaan Parikh
256
asked Jan 12 at 6:34
Ishaan ParikhIshaan Parikh
256
asked Jan 12 at 6:34
Ishaan ParikhIshaan Parikh
256
asked Jan 12 at 6:34
Ishaan ParikhIshaan Parikh
256
asked Jan 12 at 6:34
Ishaan ParikhIshaan Parikh
256
256
$begingroup$
How many equations do you consider “many?” If you can write down the equation of the circle, you can also write down the equation of the polar of a point. The points of tangency are the solutions to this system of two equations. That doesn’t sound particularly tedious to me.
$endgroup$
– amd
Jan 12 at 9:32
$begingroup$
Why do we need calculus to begin with? Analytic geometry works perfectly well.
$endgroup$
– Oscar Lanzi
Jan 12 at 11:11
add a comment |
$begingroup$
How many equations do you consider “many?” If you can write down the equation of the circle, you can also write down the equation of the polar of a point. The points of tangency are the solutions to this system of two equations. That doesn’t sound particularly tedious to me.
$endgroup$
– amd
Jan 12 at 9:32
$begingroup$
Why do we need calculus to begin with? Analytic geometry works perfectly well.
$endgroup$
– Oscar Lanzi
Jan 12 at 11:11
$begingroup$
How many equations do you consider “many?” If you can write down the equation of the circle, you can also write down the equation of the polar of a point. The points of tangency are the solutions to this system of two equations. That doesn’t sound particularly tedious to me.
$endgroup$
– amd
Jan 12 at 9:32
$begingroup$
How many equations do you consider “many?” If you can write down the equation of the circle, you can also write down the equation of the polar of a point. The points of tangency are the solutions to this system of two equations. That doesn’t sound particularly tedious to me.
$endgroup$
– amd
Jan 12 at 9:32
$begingroup$
Why do we need calculus to begin with? Analytic geometry works perfectly well.
$endgroup$
– Oscar Lanzi
Jan 12 at 11:11
$begingroup$
Why do we need calculus to begin with? Analytic geometry works perfectly well.
$endgroup$
– Oscar Lanzi
Jan 12 at 11:11
add a comment |
3 Answers
3
active
oldest
votes
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Equation of a circle is degree 2. Equation of a line passing through a given point is degree 1. So their intersection is obtained by solving a quadratic equation. For a tangent, they have only one point in commun, so the discriminant is 0...
answered Jan 12 at 6:49
amaama
335137
$endgroup$
$begingroup$
The OP asked for a solution that uses calculus.
$endgroup$
– amd
Jan 12 at 9:23
$begingroup$
No. The OP said he thought he should be using calculus. In fact you don't. This approach is a good one but I think it should be developed further anyway, to be useful.
$endgroup$
– Oscar Lanzi
Jan 12 at 11:15
add a comment |
$begingroup$
We maximize slope from a given fixed point $ (x_1,y_1)$ to a parameterized circle variable point in its standard form:
$$ (x-h) = a cos t,, (y-k)= a sin t; tag1 $$
and solve for the parameter
$$ frac{y-y_1}{x-x_1}= frac{k+ a sin t -y_1}{h+ a cos t -x_1}=frac{-cos t} {sin t}tag2$$
the latter has been obtained by chain rule differentiation
Simplifying get the condition to obtain $t_1$ of tangent points (two)
$$ sin t_1 (k-y_1)+ cos t_1 (h-x_1)+a=0 tag3 $$
It can be also recognized as tangent-normal form of a straight line.
answered Jan 12 at 7:56
NarasimhamNarasimham
20.9k62158
$endgroup$
add a comment |
$begingroup$
Let us call $R$ the ray and $C = (x_0, y_0)$ the center of the circle. The points of the circle are the $M =(Rcostheta, Rsintheta)$.
Let $E = (a, b)$ be the external point.
Then the two points $M$ corresponding to a tangeant passing through $E$ are such that vector $vec{CM}$ is orthogonal to $vec{EM}$.
It immediately gives:
$$costheta,(x_0+ Rcostheta - a)+ sintheta,(y_0+Rsintheta-b) = 0$$
And then
$$costheta,(x_0-a) + sintheta,(y_0-b) + R$$
One possibility to go further is to rewrite $vec{EC}$ as $(rhocosphi, rhosinphi )$
And therefore
$$cos(theta - phi) = -frac{R}{rho}$$
answered Jan 12 at 9:04
DamienDamien
59714
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Equation of a circle is degree 2. Equation of a line passing through a given point is degree 1. So their intersection is obtained by solving a quadratic equation. For a tangent, they have only one point in commun, so the discriminant is 0...
answered Jan 12 at 6:49
amaama
335137
$endgroup$
$begingroup$
The OP asked for a solution that uses calculus.
$endgroup$
– amd
Jan 12 at 9:23
$begingroup$
No. The OP said he thought he should be using calculus. In fact you don't. This approach is a good one but I think it should be developed further anyway, to be useful.
$endgroup$
– Oscar Lanzi
Jan 12 at 11:15
add a comment |
$begingroup$
Equation of a circle is degree 2. Equation of a line passing through a given point is degree 1. So their intersection is obtained by solving a quadratic equation. For a tangent, they have only one point in commun, so the discriminant is 0...
answered Jan 12 at 6:49
amaama
335137
$endgroup$
$begingroup$
The OP asked for a solution that uses calculus.
$endgroup$
– amd
Jan 12 at 9:23
$begingroup$
No. The OP said he thought he should be using calculus. In fact you don't. This approach is a good one but I think it should be developed further anyway, to be useful.
$endgroup$
– Oscar Lanzi
Jan 12 at 11:15
add a comment |
$begingroup$
Equation of a circle is degree 2. Equation of a line passing through a given point is degree 1. So their intersection is obtained by solving a quadratic equation. For a tangent, they have only one point in commun, so the discriminant is 0...
answered Jan 12 at 6:49
amaama
335137
$endgroup$
Equation of a circle is degree 2. Equation of a line passing through a given point is degree 1. So their intersection is obtained by solving a quadratic equation. For a tangent, they have only one point in commun, so the discriminant is 0...
answered Jan 12 at 6:49
amaama
335137
answered Jan 12 at 6:49
amaama
335137
answered Jan 12 at 6:49
amaama
335137
answered Jan 12 at 6:49
amaama
335137
335137
$begingroup$
The OP asked for a solution that uses calculus.
$endgroup$
– amd
Jan 12 at 9:23
$begingroup$
No. The OP said he thought he should be using calculus. In fact you don't. This approach is a good one but I think it should be developed further anyway, to be useful.
$endgroup$
– Oscar Lanzi
Jan 12 at 11:15
add a comment |
$begingroup$
The OP asked for a solution that uses calculus.
$endgroup$
– amd
Jan 12 at 9:23
$begingroup$
No. The OP said he thought he should be using calculus. In fact you don't. This approach is a good one but I think it should be developed further anyway, to be useful.
$endgroup$
– Oscar Lanzi
Jan 12 at 11:15
$begingroup$
The OP asked for a solution that uses calculus.
$endgroup$
– amd
Jan 12 at 9:23
$begingroup$
The OP asked for a solution that uses calculus.
$endgroup$
– amd
Jan 12 at 9:23
$begingroup$
No. The OP said he thought he should be using calculus. In fact you don't. This approach is a good one but I think it should be developed further anyway, to be useful.
$endgroup$
– Oscar Lanzi
Jan 12 at 11:15
$begingroup$
No. The OP said he thought he should be using calculus. In fact you don't. This approach is a good one but I think it should be developed further anyway, to be useful.
$endgroup$
– Oscar Lanzi
Jan 12 at 11:15
add a comment |
$begingroup$
We maximize slope from a given fixed point $ (x_1,y_1)$ to a parameterized circle variable point in its standard form:
$$ (x-h) = a cos t,, (y-k)= a sin t; tag1 $$
and solve for the parameter
$$ frac{y-y_1}{x-x_1}= frac{k+ a sin t -y_1}{h+ a cos t -x_1}=frac{-cos t} {sin t}tag2$$
the latter has been obtained by chain rule differentiation
Simplifying get the condition to obtain $t_1$ of tangent points (two)
$$ sin t_1 (k-y_1)+ cos t_1 (h-x_1)+a=0 tag3 $$
It can be also recognized as tangent-normal form of a straight line.
answered Jan 12 at 7:56
NarasimhamNarasimham
20.9k62158
$endgroup$
add a comment |
$begingroup$
We maximize slope from a given fixed point $ (x_1,y_1)$ to a parameterized circle variable point in its standard form:
$$ (x-h) = a cos t,, (y-k)= a sin t; tag1 $$
and solve for the parameter
$$ frac{y-y_1}{x-x_1}= frac{k+ a sin t -y_1}{h+ a cos t -x_1}=frac{-cos t} {sin t}tag2$$
the latter has been obtained by chain rule differentiation
Simplifying get the condition to obtain $t_1$ of tangent points (two)
$$ sin t_1 (k-y_1)+ cos t_1 (h-x_1)+a=0 tag3 $$
It can be also recognized as tangent-normal form of a straight line.
answered Jan 12 at 7:56
NarasimhamNarasimham
20.9k62158
$endgroup$
add a comment |
$begingroup$
We maximize slope from a given fixed point $ (x_1,y_1)$ to a parameterized circle variable point in its standard form:
$$ (x-h) = a cos t,, (y-k)= a sin t; tag1 $$
and solve for the parameter
$$ frac{y-y_1}{x-x_1}= frac{k+ a sin t -y_1}{h+ a cos t -x_1}=frac{-cos t} {sin t}tag2$$
the latter has been obtained by chain rule differentiation
Simplifying get the condition to obtain $t_1$ of tangent points (two)
$$ sin t_1 (k-y_1)+ cos t_1 (h-x_1)+a=0 tag3 $$
It can be also recognized as tangent-normal form of a straight line.
answered Jan 12 at 7:56
NarasimhamNarasimham
20.9k62158
$endgroup$
We maximize slope from a given fixed point $ (x_1,y_1)$ to a parameterized circle variable point in its standard form:
$$ (x-h) = a cos t,, (y-k)= a sin t; tag1 $$
and solve for the parameter
$$ frac{y-y_1}{x-x_1}= frac{k+ a sin t -y_1}{h+ a cos t -x_1}=frac{-cos t} {sin t}tag2$$
the latter has been obtained by chain rule differentiation
Simplifying get the condition to obtain $t_1$ of tangent points (two)
$$ sin t_1 (k-y_1)+ cos t_1 (h-x_1)+a=0 tag3 $$
It can be also recognized as tangent-normal form of a straight line.
answered Jan 12 at 7:56
NarasimhamNarasimham
20.9k62158
edited Jan 12 at 8:05
answered Jan 12 at 7:56
NarasimhamNarasimham
20.9k62158
answered Jan 12 at 7:56
NarasimhamNarasimham
20.9k62158
answered Jan 12 at 7:56
NarasimhamNarasimham
20.9k62158
20.9k62158
add a comment |
add a comment |
$begingroup$
Let us call $R$ the ray and $C = (x_0, y_0)$ the center of the circle. The points of the circle are the $M =(Rcostheta, Rsintheta)$.
Let $E = (a, b)$ be the external point.
Then the two points $M$ corresponding to a tangeant passing through $E$ are such that vector $vec{CM}$ is orthogonal to $vec{EM}$.
It immediately gives:
$$costheta,(x_0+ Rcostheta - a)+ sintheta,(y_0+Rsintheta-b) = 0$$
And then
$$costheta,(x_0-a) + sintheta,(y_0-b) + R$$
One possibility to go further is to rewrite $vec{EC}$ as $(rhocosphi, rhosinphi )$
And therefore
$$cos(theta - phi) = -frac{R}{rho}$$
answered Jan 12 at 9:04
DamienDamien
59714
$endgroup$
add a comment |
$begingroup$
Let us call $R$ the ray and $C = (x_0, y_0)$ the center of the circle. The points of the circle are the $M =(Rcostheta, Rsintheta)$.
Let $E = (a, b)$ be the external point.
Then the two points $M$ corresponding to a tangeant passing through $E$ are such that vector $vec{CM}$ is orthogonal to $vec{EM}$.
It immediately gives:
$$costheta,(x_0+ Rcostheta - a)+ sintheta,(y_0+Rsintheta-b) = 0$$
And then
$$costheta,(x_0-a) + sintheta,(y_0-b) + R$$
One possibility to go further is to rewrite $vec{EC}$ as $(rhocosphi, rhosinphi )$
And therefore
$$cos(theta - phi) = -frac{R}{rho}$$
answered Jan 12 at 9:04
DamienDamien
59714
$endgroup$
add a comment |
$begingroup$
Let us call $R$ the ray and $C = (x_0, y_0)$ the center of the circle. The points of the circle are the $M =(Rcostheta, Rsintheta)$.
Let $E = (a, b)$ be the external point.
Then the two points $M$ corresponding to a tangeant passing through $E$ are such that vector $vec{CM}$ is orthogonal to $vec{EM}$.
It immediately gives:
$$costheta,(x_0+ Rcostheta - a)+ sintheta,(y_0+Rsintheta-b) = 0$$
And then
$$costheta,(x_0-a) + sintheta,(y_0-b) + R$$
One possibility to go further is to rewrite $vec{EC}$ as $(rhocosphi, rhosinphi )$
And therefore
$$cos(theta - phi) = -frac{R}{rho}$$
answered Jan 12 at 9:04
DamienDamien
59714
$endgroup$
Let us call $R$ the ray and $C = (x_0, y_0)$ the center of the circle. The points of the circle are the $M =(Rcostheta, Rsintheta)$.
Let $E = (a, b)$ be the external point.
Then the two points $M$ corresponding to a tangeant passing through $E$ are such that vector $vec{CM}$ is orthogonal to $vec{EM}$.
It immediately gives:
$$costheta,(x_0+ Rcostheta - a)+ sintheta,(y_0+Rsintheta-b) = 0$$
And then
$$costheta,(x_0-a) + sintheta,(y_0-b) + R$$
One possibility to go further is to rewrite $vec{EC}$ as $(rhocosphi, rhosinphi )$
And therefore
$$cos(theta - phi) = -frac{R}{rho}$$
answered Jan 12 at 9:04
DamienDamien
59714
edited Jan 12 at 9:15
answered Jan 12 at 9:04
DamienDamien
59714
answered Jan 12 at 9:04
DamienDamien
59714
answered Jan 12 at 9:04
DamienDamien
59714
59714
add a comment |
add a comment |
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$begingroup$
How many equations do you consider “many?” If you can write down the equation of the circle, you can also write down the equation of the polar of a point. The points of tangency are the solutions to this system of two equations. That doesn’t sound particularly tedious to me.
$endgroup$
– amd
Jan 12 at 9:32
$begingroup$
Why do we need calculus to begin with? Analytic geometry works perfectly well.
$endgroup$
– Oscar Lanzi
Jan 12 at 11:11