# Homework Help: Line, Circle, Locus problem

1. Apr 27, 2015

### Raghav Gupta

1. The problem statement, all variables and given/known data
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

2. Relevant equations
NA
3. 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?

Last edited: Apr 27, 2015
2. Apr 27, 2015

### 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 ?

3. Apr 27, 2015

### Raghav Gupta

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?

4. Apr 27, 2015

### BvU

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

5. Apr 27, 2015

### Raghav Gupta

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

6. Apr 27, 2015

### 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 ?

7. Apr 27, 2015

### Raghav Gupta

Oh, so you are seeing my attempt?

8. Apr 27, 2015

### 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.

Last edited: Apr 27, 2015
9. Apr 27, 2015

### Raghav Gupta

Yeah we can agree.
Now how to proceed?

10. Apr 27, 2015

### BvU

11. Apr 27, 2015

### HallsofIvy

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

12. Apr 27, 2015

### BvU

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 !

13. Apr 27, 2015

### Raghav Gupta

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

14. Apr 27, 2015

### 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 !)​

Last edited: Apr 27, 2015
15. Apr 28, 2015

### Raghav Gupta

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.

16. Apr 28, 2015

### 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)

17. Apr 28, 2015

### Raghav Gupta

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 ?

18. Apr 28, 2015

### 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 ?

--

19. Apr 28, 2015

### Raghav Gupta

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

20. Apr 28, 2015

### 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 $$x\cos\alpha + y\sin\alpha = |n| \quad {\rm or} \quad \vec v \cdot \hat n = |\vec n|$$
--

Last edited: Apr 28, 2015
21. Apr 28, 2015

### Raghav Gupta

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

Last edited: Apr 28, 2015
22. Apr 28, 2015

### Raghav Gupta

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

23. Apr 28, 2015

### Ray Vickson

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

24. Apr 28, 2015

### BvU

Thank you for the compliment; that was MS Visio.

25. Apr 29, 2015

### Raghav Gupta

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