What is the Locus of Points on a Circle Touched by a Given Line?

[Raghav Gupta]

Homework Statement

If the line $hk + ky = 1$ touches the circle $x^2 + y^2 = a^2$, then locus of the points (h,k) is a circle of radius :
a
h
k
1/a

Homework Equations

NA

The Attempt at a Solution

Tried differentiating circle equation (Is differentiation of use here?, also don't know if it is a calculus or pre-calculus problem. )
Got y' = -x/y
Differentiated line equation and thinking h,k are constants which is not true I got y' = 0
I have to substitute something but what?

 

[BvU]

Make a drawing !
You know the slope of the line, you know an equation for a point it has to go through.

And you probably have a typo in the equation for the line ?

 

[Raghav Gupta]

Make a drawing !
You know the slope of the line, you know an equation for a point it has to go through.

And you probably have a typo in the equation for the line ?

No, I have not a typo in equation of line.
The question is confusing me as I think there should be x in equation of line?

 

[BvU]

Now it says $y={1\over k}-h$ having slope zero. Touches the circle if ...

 

[Raghav Gupta]

Now it says $y={1\over k}-h$ having slope zero. Touches the circle if ...

Not getting, are you saying I have a typo in Question statement of line equation?

 

[BvU]

That is what I said and that is what I typed. $hk + ky = 1$ is the equation for a horizontal line. Slope 0. Does it say that in the exercise you were given ?

 

[Raghav Gupta]

That is what I said and that is what I typed. $hk + ky = 1$ is the equation for a horizontal line. Slope 0. Does it say that in the exercise you were given ?

Oh, so you are seeing my attempt?

The Attempt at a Solution

Differentiated line equation and thinking h,k are constants which is not true I think I got y' = 0
I have to substitute something but what?

The exercise says only the problem statement.

 

[BvU]

Can we please agree that the exercise text states $hk + ky = 1$ as the equation of the line ? If so, we can proceeed and prove that the locus of the points (h,k) is not a circle.

 

[Raghav Gupta]

Can we please agree that the exercise tekst states $hk + ky = 1$ as the equation of the line ? If so, we can proceeed and prove that the locus of the points (h,k) is not a circle.

Yeah we can agree.
Now how to proceed?

 

[BvU]

Now it says $y={1\over k}-h$ having slope zero. Touches the circle if ...

 

[HallsofIvy]

Is it hk+ ky= 1 or hx+ ky= 1?

 

[BvU]

Is it hk+ ky= 1 or hx+ ky= 1?

In post #9 Raghav repeats hk+ ky= 1 and now we are under way to discover that that does NOT give a circle as locus. Read !

 

[Raghav Gupta]

Now it says $y={1\over k}-h$ having slope zero. Touches the circle if ...

y = a or y = -a.
So the question is wrong as locus is not a circle but mere two points.

 

[BvU]

Not the way it works. You are asked for the locus of the points (h,k), not for the points (x,y) where the line touches the circles.

To make a long story short: $y=\pm a$ is correct as the equation for the tangent line. If you want to write that as $hk + ky = 1$ you get e.g. $hk + ka = 1$ so the points (h,k) satisfy $k = {1\over h+a}$ which is not the equation of a circle.

Having delivered proof that $hk + ky = 1$ is not what was intended for this exercise, I propose you now accept that it should have been $hx + ky = 1$.

What do we know about these lines ?

Well, as you already found in post #1, they have slope -x/y . Also, one of the points on the line satisfies $x^2+y^2 = a^2$. So what do we have to impose on h and k to make the equation for such a line take the form $hx + ky = 1$ ?

(My advice: draw a circle and a tangent !)​

 

[Raghav Gupta]

Well, as you already found in post #1, they have slope -x/y . Also, one of the points on the line satisfies $x^2+y^2 = a^2$. So what do we have to impose on h and k to make the equation for such a line take the form $hx + ky = 1$ ?

(My advice: draw a circle and a tangent !)​

Took a point on line x1,y1
Then slope is - x1/y1.
x1² + y1² = a²
Also hx1 + ky1 = 1.
Here is a circle of radius a and some random 2 tangents drawn.

 

[BvU]

Good start. The point where the line is tangent to the circle is (x,1y1).
Now (important step): what's the equation of the line ? (I don't mean hx1 + ky1 = 1 but the equation for the line in therms of x1, y1 and a)

 

[Raghav Gupta]

Good start. The point where the line is tangent to the circle is (x,1y1).
Now (important step): what's the equation of the line ? (I don't mean hx1 + ky1 = 1 but the equation for the line in therms of x1, y1 and a)

I was doing some manipulations,
Take the general equation

$hx + ky = 1$

$⇒ y = \frac{1-hx}{k}$

In equation $x^2 + y^2 = 1$

Substituting y value as it satisfies,

$x^2 + \frac{1+ h^2x^2 -2hx}{k^2} = a^2$
Rearranging,

$x^2( k^2 + h^2) - 2hx +1 - k^2a^2 = 0$

It striked to me that

Now discriminant of quadratic equation should be zero as tangent line touches one point.

$4h^2 - 4 (k^2 + h^2)(1-k^2a^2) = 0$

Solving we get,

$k^2 + h^2 = 1/a^2$
So locus is circle of radius 1/a

What was the use of diagram here and calculating slope earlier as -x/y ?

 

[BvU]

LIne segment from origin to (x1, y1) has length a.
Tangent line in (x1, y1) is perpendicular to that line.

That means all points (x,y) on that tangent line satisfy $$ (x_1,y_1)\cdot(x, y) = a^2$$ -- an inner product, written out : $$x_1 x + y_1 y = a^2$$ -- and there's your tangent line equation without any quadratic equation at all !
Compare $$x_1 x + y_1 y = a^2 \quad {\rm to} \quad x + k y = 1$$ and you see
$$h = x_1/a^2 \quad {\rm and} \quad k = y_1/a^2$$
Since $$ x_1^2+ y_1^2 =a^2$$ you indeed find
$$ h^2+ k^2 =1/a^2$$

Well done !

I hope you are now also convinced that the line equation in post #1 was a typo ?

--

 

[Raghav Gupta]

Oh, another and easy way

LIne segment from origin to (x1, y1) has length a.
Tangent line in (x1, y1) is perpendicular to that line.

That means all points (x,y) on that tangent line satisfy $$ (x_1,y_1)\cdot(x, y) = a^2$$ -- an inner product, written out : $$x_1 x + y_1 y = a^2$$ -- and there's your tangent line equation without any quadratic equation at all !--

Not getting this
$$ (x_1,y_1)\cdot(x, y) = a^2$$ -- an inner product,

 

[BvU]

Have you ever heard of the normal equation for a straight line ? We learned it at school (I was 15 then...).
It goes like this:

For all points on the dashed tangent: Inner product $$(x,y) \cdot\ (x_1, y_1) = \vec v_1 \cdot \vec v_2 = |\vec v_1 |\;|\vec v_2| \;\cos\phi = |\vec v_1 |\;|\vec v_1|/ \cos\phi \;\cos\phi = |\vec v_1 |^2$$
Note that this time the arrows aren't for infinite lines but for vectors.

There is also a unit normal vector form: if $\tan \alpha = y_1/x_1$ (hey, that's like (-x_1/y_1) ^-1 !)

you have with $\vec n = (x_1, y_1), \quad \hat n = (\cos\alpha, \sin\alpha) = (x_1, y_1) / |\vec v_1|$ that $$ x\cos\alpha + y\sin\alpha = |n| \quad {\rm or} \quad \vec v \cdot \hat n = |\vec n|$$
--

 

[Raghav Gupta]

I hope you are now also convinced that the line equation in post #1 was a typo ?

--

Yeah, ( The typo was by the book ,not by me)

 

[Raghav Gupta]

Have you ever heard of the normal equation for a straight line ? We learned it at school (I was 15 then...).
It goes like this:

View attachment 82697
For all points on the dashed tangent: Inner product $$(x,y) \cdot\ (x_1, y_1) = \vec v_1 \cdot \vec v_2 = |\vec v_1 |\;|\vec v_2| \;\cos\phi = |\vec v_1 |\;|\vec v_1|/ \cos\phi \;\cos\phi = |\vec v_1 |^2$$
Note that this time the arrows aren't for infinite lines but for vectors.

There is also a unit normal vector form: if $\tan \alpha = y_1/x_1$ (hey, that's like (-x_1/y_1) ^-1 !)

you have with $\vec n = (x_1, y_1), \quad \hat n = (\cos\alpha, \sin\alpha) = (x_1, y_1) / |\vec v_1|$ that $$ x\cos\alpha + y\sin\alpha = |n| \quad {\rm or} \quad \vec v \cdot \hat n = |\vec n|$$
--

I have not read that form earlier but now see it is fast as compared to quadratic.
Thanks.

 

[Ray Vickson]

Have you ever heard of the normal equation for a straight line ? We learned it at school (I was 15 then...).
It goes like this:

View attachment 82697
For all points on the dashed tangent: Inner product $$(x,y) \cdot\ (x_1, y_1) = \vec v_1 \cdot \vec v_2 = |\vec v_1 |\;|\vec v_2| \;\cos\phi = |\vec v_1 |\;|\vec v_1|/ \cos\phi \;\cos\phi = |\vec v_1 |^2$$
Note that this time the arrows aren't for infinite lines but for vectors.

There is also a unit normal vector form: if $\tan \alpha = y_1/x_1$ (hey, that's like (-x_1/y_1) ^-1 !)

you have with $\vec n = (x_1, y_1), \quad \hat n = (\cos\alpha, \sin\alpha) = (x_1, y_1) / |\vec v_1|$ that $$ x\cos\alpha + y\sin\alpha = |n| \quad {\rm or} \quad \vec v \cdot \hat n = |\vec n|$$
--

Very nice drawing! What software package did you use to draw it?

 

[BvU]

Thank you for the compliment; that was MS Visio.

 

[Raghav Gupta]

Have you ever heard of the normal equation for a straight line ? We learned it at school (I was 15 then...).
It goes like this:

View attachment 82697
For all points on the dashed tangent: Inner product $$(x,y) \cdot\ (x_1, y_1) = \vec v_1 \cdot \vec v_2 = |\vec v_1 |\;|\vec v_2| \;\cos\phi = |\vec v_1 |\;|\vec v_1|/ \cos\phi \;\cos\phi = |\vec v_1 |^2$$
Note that this time the arrows aren't for infinite lines but for vectors.

There is also a unit normal vector form: if $\tan \alpha = y_1/x_1$ (hey, that's like (-x_1/y_1) ^-1 !)

you have with $\vec n = (x_1, y_1), \quad \hat n = (\cos\alpha, \sin\alpha) = (x_1, y_1) / |\vec v_1|$ that $$ x\cos\alpha + y\sin\alpha = |n| \quad {\rm or} \quad \vec v \cdot \hat n = |\vec n|$$
--

Hey what angle $\alpha$ is here in diagram?
I understood your dashed tangent inner product but not that unit vector normal form.

 

[BvU]

$\tan\alpha = y_1/x_1$

This unit vector normal form simply removes one factor $|\vec n|$.