# Another Locus Problem

1. Apr 28, 2015

### Raghav Gupta

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

If the tangent from a point P to the circle $x^2 + y^2 = 1$ is perpendicular to the tangent from P to the circle $x^2 + y^2 = 3$, then the locus of P is:
2. Relevant equations
Equation of tangent to a circle in terms of slope m is,
$y = mx ± a\sqrt{1+ m^2}$

3. The attempt at a solution
See diagram,

I know $m_1m_2 = -1$
getting $y = m_1x ± \sqrt{ 1+ m_1^2}$
and other equation $y = m_2x ± √3\sqrt{1+ m_2^2}$
I think I need one more equation for getting locus equation but from where it will come?

2. Apr 28, 2015

### SammyS

Staff Emeritus
What is the radius of this circle?

Where is the center of this circle?

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

### Raghav Gupta

The two circles are concentric having centre( 0,0) and first circle has radius of 1 and other circle has radius of $\sqrt{3}$

4. Apr 28, 2015

### BvU

Dear Raghav,

If you would just make a slightly better drawing... your 90 degrees doesn't look like 90 degrees at all !

And "locus" means: you have to find all points that have this property, so move one tangent and let the other follow. The answer will hit you in the face !

5. Apr 28, 2015

### SammyS

Staff Emeritus
Yes, quite clearly.

I tried to indicate by how I separated the quotations --

Under 2. Relevant equations,

What is the radius, and location of the center of the circle that the following is true for?

Equation of tangent to a circle in terms of slope m is,
$\displaystyle\ y = mx ± a\sqrt{1+ m^2}\$​

Is the point, $\ (x,\,y)\,,\$ located on the circle or located on the tangent line?

When you go to use some derived quantity, you need to know under what conditions it is valid.

6. Apr 28, 2015

### Raghav Gupta

I drawed once more but the answers are not hitting my face.
I think one should know the art of drawing to solve these problems.
Are you using some software for making drawings?
That would be cheating, since I am drawing by hand.

7. Apr 28, 2015

### Raghav Gupta

The (x,y) is located on tangent line.
That formula is valid when any line touches the circle at one point.

8. Apr 28, 2015

### BvU

I don't call that cheating. But you can also draw two circles using a cup and a saucer. And a 90 degree angle is available from a piece of cardboard. Are we going to argue about this ?

You can always use the whiteboard from the button on the lower right to make a point

9. Apr 28, 2015

### Raghav Gupta

I don't use a computer and PF whiteboard is not applicable for mobile devices. ( The PF whiteboard was a recent addition in response to my thread in feedback and announcement forums. )
By diagram, it is looking the locus of P is a circle but of what radius?
Now I think apart from drawing we will definitely need some algebra for solving this problem?

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

### SammyS

Staff Emeritus
(That diagram is so informative, that you should be able to determine the radius by using it.)

To do this by algebra. ...

If the two tangent lines are to be perpendicular, how are m1 and m2 related?

11. Apr 28, 2015

### BvU

Algebra: you recently got acquainted with the normal equation for a tangent line . Very useful here!

Geometry: no computer, so back to cup, saucer and cardboard (but instead of the cardoard you can use a phone, I suppose -- or is that cheating ?).
Anyway, if you draw these normal vectors in the figure (or imagine them), you have three 90 degree angles in a quadrangle

12. Apr 28, 2015

### Ray Vickson

Back when I was a student we used rulers, compasses and protractors to make drawings. We purchased graph paper for making plots (although these days you can download and print your own), using a gadget called a "French curve" to make smooth graphs through some points. I am not sure you can even find a French curve anymore, but the other items are still available. As has been suggested, you can use a glass or a cup for drawing circles and the frame of an i-phone to make right-angles.

13. Apr 29, 2015

### Raghav Gupta

Thanks to all of you three guys.
Very true. Applied Pythagoras theorem and got radius 2.
I think diagram was more useful. Centre of circle is (0,0). So taking a point (x,y). Applying distance formula
$x^2 + y^2 = 4$ That distance 2 was found from diagram.
Nah, I think diagram and Pythagoras was more useful.
Phone is a cheating, because we cannot bring phone in an examination hall.
The inviligators would consider us mad if we bring cup, saucer and cardboard in an examination hall.
Your idea of protractor, graphs , ruler is good. But I think rough drawing is a fast way.
And I do not have an i- phone.

Last edited: Apr 29, 2015
14. Apr 29, 2015

### BvU

Well well, are you now changing course altogether and dropping the algebraic approach in favour of the geometrical ?
If I were you I would at least try to come to the same result with the more abstract algebraic approach (*). Seeing it in a picture is for pussies, a real scientist comes to results with formulas !

(*) great advantage:more extendable to really complex problems in many more dimensions

15. Apr 29, 2015

### Raghav Gupta

I am getting, by that inner product form for tangents,
xx1 + yy1 = 1
xx2 + yy2 = 3
Also since tangents are perpendicular at P
x1x2/(y1y2) = -1

16. Apr 29, 2015

### Raghav Gupta

m1m2 = -1 .

17. Apr 29, 2015

### certainly

These are still available, they're required in courses on Technical Drawing ..............

18. Apr 29, 2015

### BvU

Good. You also know $(x_1,y_1)$ can be written as $(\cos\alpha, \sin\alpha)$ and
$(x_2,y_2)$ can be written as $\sqrt 3 (\cos\beta, \sin\beta)=\sqrt 3(\cos(\alpha+\pi/2), \sin(\alpha+\pi/2) = \sqrt 3(- \sin\alpha, \cos\alpha)$.

And you know point P is on both lines. 2 eqns in 2 unknowns (x,y) with one parameter, $\alpha$.

--

19. Apr 29, 2015

### Raghav Gupta

How β = α + π/2 ?

20. Apr 29, 2015

### BvU

Because the lines are perpendicular the difference in angle is 90 degrees, aka $\pi/2$