Finding 2 Tangents Of A Circle Given A Point?

kei10 · 2011-01-20 19:00:42

Hi, I'm new to this forum.

Okay here's my question was...
Is it possible to find the point of the tangent of a circle given just a point?
There will be 2 tangents unless the point is directly on the circumference.

I know it's possible with given an equation of the circle and an equation of a slope.
But I can't solve this one...

Solving it without the use of Graph but just formula.
Given the center of the circle, (0,0) with the radius = 5.
And a point, (7,1).
The equation of the circle will be; [x^2 + y^2 = 25]
While the equation of the slope is unknown.

We have to find the point of the intersection of (x1,y1) and (x2,y2).
Is this possible? If there is, how?

Thanks.

Last edited by kei10 (2011-01-20 19:05:42)

Raghuram · 2011-01-20 20:58:35

It can be solved by using properties of Tanzent.

The tanzent and radius intersect at right angles. Hence, the product of slope of tanzent and radius is equal to -1.

The slope of tanzent is (Y1-1)/(x1-7) and slope of radius is (y1-0)/(X1-0). The above property yeilds to an equation:

((Y1-1)*Y1)/((X1-7)*X1)=-1 ---- Y1=25-7(X1) -------- 1

Radius of circle is 25 units, this yields to x1^2+Y1^2 = 25 ------- 2

By solving above two equations we can get the coordinates ----- Answer is (3,4) & (4,-3)

We get two points because one of the equation is a Quadratic equation.

bobbym · 2011-01-20 22:23:23

Hi kei10 and Raghuram;

Welcome to the forum! Here is another way that uses more algebra.

This is the general equation that passes through the point (7,1)

Where m is the slope.

Substituting 2 into 1:

Expanding and cleaning up

We have an equation of the points that intersect 1) and 2). We want 1 intersection ( 1 root of multiplicity 2 ), this will be the tangents we require. To get 1 root we set the discriminant = 0. That will find the m that gives 3) one root of multiplicity 2.

The discriminant of a quadratic is b^2 - 4ac

Only looks tough but it is not. Simplifying and setting to 0 we get.

Solving for m

Now we have the 2 slopes we require.

The two tangents are:

kei10 · 2011-01-21 02:18:43

I've never thought of using discriminant of quadratic, I'm dumb!

But...

Oh boy oh boy thank you!!
Unbelievably splendidly awesome!!
Thank you very much.
.. Thanks for a dozen more than a bunch!

Thanks.

bobbym · 2011-01-21 02:25:33

Hi kei10;

Your welcome!

soroban · 2011-01-21 03:37:54

. .

. . . .

.

kei10 · 2011-01-21 11:30:33

soroban wrote:

. .

. . . .

.

Thanks, It helps a lot.
But in other words, that probably works only when the shape of the tangent is perpendicular.
But bobbym's method works on every points, and even tangents.
... except it needs a lot of hard work on expanding/cleaning up the equations.

Thanks a lot more than a bunch!

bobbym · 2011-01-21 11:36:30

Hi kei10;

That is not exactly correct, the tangent is always perpendicular to the radius at the point. Unless I am misunderstanding it their method will always work too.

Please understand that I only provided the method I used because Raghuram had already solved it in that way. I just wanted to show you that there is always another way.

Soroban is a very clever guy and it pays to look over his ideas.

You can of course use the method that you are most comfortable with. That is a smart way to work.

kei10 · 2011-01-21 11:54:41

bobbym wrote:

Hi kei10;
That is not exactly correct, the tangent is always perpendicular to the radius at the point. Unless I am misunderstanding it their method will always work too.
Please understand that I only provided the method I used because Raghuram had already solved it in that way. I just wanted to show you that there is always another way.
Soroban is a very clever guy and it pays to look over his ideas.
You can of course use the method that you are most comfortable with. That is a smart way to work.

Hi.

Perhaps I'm wrongly analyzed the method.
I think I am having difficulty of your method since it's so hard.
Then I will try comparing to sorobans and raghuram's.
I'll try harder anyway.
Well, Thanks!

Last edited by kei10 (2011-01-21 12:04:03)

bobbym · 2011-01-21 12:01:34

It is a fairly tough problem. Yes, there is a lot of tedious algebra in it.

kei10 · 2011-01-21 17:25:16

I manage to solve it for the question below.

Sorry I don't know how to use the [math.][/math.] thing.

I tried with your method:
⇒ [y = m(x-10)+1] ⇒ x² + y² = 25
⇒ x² + (mx - 10m + 1)² - 25 = 0 ...... Expand and cleanup...
⇒ m²x² - 20m²x + 100m² +2mx - 20m + x² - 24 = 0 ...... Into quadratic equation.
⇒ (m² + 1)x² + (2m - 20m²)x + 100m² - 20m - 24 = 0 ...... Into The discriminant of a quadratic.
⇒ (2m - 20m²)² - 4(m² + 1)(100m² - 20m - 24) = 0 ...... Poof...!
⇒ 300m² - 80m - 96 = 0 ...... Solve with (-b±√(b² - 4ac))/(2a)...
⇒ m1 = (2+2√19)/15 and m2 = (2-2√19)/15 ...... Now we have the gradient, c = y - mx...
⇒ c1 = 1 - 10m1 and c2 = 1 - 10m2 ...... So...
⇒ c1 = (-1-4√19)/3 and c2 = (-1+4√19)/3 ....Subtitute into x² + y² = 25...
⇒ x² + (mx + c)² - 25 = 0 ...... Expand and cleanup...
⇒ (m² + 1)x² + 2mcx + c² - 25 = 0 ...... Solve with (-b±√(b² - 4ac))/(2a)...
⇒ (-(2mc) + √((2mc)² - 4(m² + 1)(c² - 25)))/(2(m² + 1)) ... THIS IS MADNESS...
⇒ x1 = 2.906821678 ...... Subtitude into y = mx + c and solve...
⇒ y1 = -4.068216776

So the first point of tangent is... approximately (2.9,-4.1)...
... There's no way I am gonna find the other point of tangent.
This is crazy. ( Took me 3 hours to solve the mistakes ).
At last I end up using online tools.

Okay, next, I tried using soroban's:
[math]⇒ (y - 1)/(x - 10) = -x/y ⇒
⇒ y² - y = -x² + 10x ⇒ 10x + y = x² + y² ⇒ y = 25 - 10x
⇒ x² + (25 - 10x)² = 25
⇒ x² + 625 - 500x + 100x² - 25 = 0 ⇒ 101x² - 500x + 600 = 0
Solving 101x² - 500x + 600 = 0 with (-b±√(b² - 4ac))/(2a)...
⇒ x1 = 2.906821678 x2 = 2.043673372 ... Am I seeing things?
Subtituting x1 and x2 into y = 25 - 10x...
⇒ y1 = -4.068216776 and y2 = 4.563266281

Resulting: (Sorry The image is over-sized...)

... x1 ≈ 2.906821678, y1 = -4.068216776 and x2 = 2.043673372, y2 = 4.563266281.
Wow... I solved this at once using soroban's method.
I'm sorry I misread the method, I thought it was something else.

Thank you Bobbym, Uh... Soroban, and Raghuram.
Thanks for a lot more than a dozen even more than a lot of a bunch!
Thank you! ( I don't know what else to say. )

Last edited by kei10 (2011-01-21 17:28:39)

bobbym · 2011-01-21 17:49:22

Hi kei10;

I do not know why you had problems with this method for ( 10 , 1 ). But use the method you like and find easier. Just want to show you that it does get there.

For your new problem with point ( 10 , 1 ) and the same circle.

substitute 2 into 1:

Take the discriminant. Set it to 0 to make sure there is only only intersection between the line and the circle.

Those are the 2 slopes use y = mx + b to get the equations.

These are the equations of the tangents.

kei10 · 2011-01-21 18:07:08

Hm... I thought I already solved using soroban's method, it's way even better and easier.
But thanks for showing it, too.
I guess you didn't read the whole post.
Maybe it's just confusing.

Anyway, Thanks.

Last edited by kei10 (2011-01-21 18:08:10)

bobbym · 2011-01-21 18:13:12

No problem. I like solving problems.

Guest12345 · 2011-01-22 01:16:52

Bobbym's complex algebraic method will able to find the slope/gradient of the tangency directly.
Whilst Raghuram/Soroban('s) will able to find the points of the tangency directly.

So, both of them are very useful.
Cheers!

bobbym · 2011-01-22 08:00:05

Hi Guest12345;

Welcome to the forum. The algebraic way can be used on other curves not just a circle. I have used it now and find that I like it. Only one thing, it is not my method. I had it in my notes which means I got it from somewhere.

Math Is Fun Forum

#1 2011-01-20 19:00:42

Finding 2 Tangents Of A Circle Given A Point?

#2 2011-01-20 20:58:35

Re: Finding 2 Tangents Of A Circle Given A Point?

#3 2011-01-20 22:23:23

Re: Finding 2 Tangents Of A Circle Given A Point?

#4 2011-01-21 02:18:43

Re: Finding 2 Tangents Of A Circle Given A Point?

#5 2011-01-21 02:25:33

Re: Finding 2 Tangents Of A Circle Given A Point?

#6 2011-01-21 03:37:54

Re: Finding 2 Tangents Of A Circle Given A Point?

#7 2011-01-21 11:30:33

Re: Finding 2 Tangents Of A Circle Given A Point?

#8 2011-01-21 11:36:30

Re: Finding 2 Tangents Of A Circle Given A Point?

#9 2011-01-21 11:54:41

Re: Finding 2 Tangents Of A Circle Given A Point?

#10 2011-01-21 12:01:34

Re: Finding 2 Tangents Of A Circle Given A Point?

#11 2011-01-21 17:25:16

Re: Finding 2 Tangents Of A Circle Given A Point?

#12 2011-01-21 17:49:22

Re: Finding 2 Tangents Of A Circle Given A Point?

#13 2011-01-21 18:07:08

Re: Finding 2 Tangents Of A Circle Given A Point?

#14 2011-01-21 18:13:12

Re: Finding 2 Tangents Of A Circle Given A Point?

#15 2011-01-22 01:16:52

Re: Finding 2 Tangents Of A Circle Given A Point?

#16 2011-01-22 08:00:05

Re: Finding 2 Tangents Of A Circle Given A Point?

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