Circle Tangents and Locus Problem: Finding the Equation with Given Slopes

[Raghav Gupta]

Homework Statement

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:

Homework Equations

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

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?

 

[SammyS]

Homework Statement

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:

Homework Equations

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

What is the radius of this circle?

Where is the center of this circle?

The Attempt at a Solution

See diagram,
View attachment 82703
I know $m_1m_2 = -1$
getting $y = m_1x ± \sqrt{ 1+ m_1^2}$
and other equation $y = m_2x ± \sqrt{3}\,\sqrt{1+ m_2^2}$
I think I need one more equation for getting locus equation but from where it will come?

 

[Raghav Gupta]

What is the radius of this circle?

Where is the center of this circle?

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

 

[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 !

 

[SammyS]

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

Yes, quite clearly.

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

Under

Homework 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.

 

[Raghav Gupta]

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 !

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.

 

[Raghav Gupta]

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.

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

 

[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

 

[Raghav Gupta]

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

View attachment 82707

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?

 

[SammyS]

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?

(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?

 

[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

 

[Ray Vickson]

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.

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.

 

[Raghav Gupta]

Thanks to all of you three guys.

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

Very true. Applied Pythagoras theorem and got radius 2.

To do this by algebra. ...

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

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.

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

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.

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.

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.

 

[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

 

[Raghav Gupta]

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

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

 

[Raghav Gupta]

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

m₁m₂ = -1 .

 

[certainly]

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,

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

 

[BvU]

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

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, $\alpha$.

And you already know the answer for x,y, right ?

--

 

[Raghav Gupta]

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, $\alpha$.

And you already know the answer for x,y, right ?

--

How β = α + π/2 ?

 

[BvU]

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

 

[Raghav Gupta]

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

Okay,
Squared the equations and added them and got the answer. Thanks.