Point of tangency to a circle from a point not on the circle

[purplecows]

Homework Statement

Find the points of tangency to a circle given by x^2+y^2=9 from point (12,9).

Homework Equations

dy/dx=-x/y
(what I've been able to come up so far)

The Attempt at a Solution

Taking the derivative I got dy/dx=-x/y

Let the unknown point of tangency be (a,b)

y-b=(-a/b)(x-a)
Simplifying that, I got:
by-ax=a^2+b^2
a and b fall on the circle; the circle's equation is x^2+y^2=9; therefore, a^2+b^2=9

by-ax=9
(12,9) is a point on this ^ line, so

9b-12a=9
b=(4/3)a+1

Substituting back into the original equation x^2+y^2=9,

a^2+((4/3)a+1)^2=9

Simplifying that got me 25a^2+27a-72=0.

This was the point where I knew I was wrong. Where did I go wrong/how do I fix it?

 

[berkeman]

Homework Statement

Find the points of tangency to a circle given by x^2+y^2=9 from point (12,9).

Homework Equations

dy/dx=-x/y
(what I've been able to come up so far)

The Attempt at a Solution

Taking the derivative I got dy/dx=-x/y

Let the unknown point of tangency be (a,b)

y-b=(-a/b)(x-a)
Simplifying that, I got:
by-ax=a^2+b^2
a and b fall on the circle; the circle's equation is x^2+y^2=9; therefore, a^2+b^2=9

by-ax=9
(12,9) is a point on this ^ line, so

9b-12a=9
b=(4/3)a+1

Substituting back into the original equation x^2+y^2=9,

a^2+((4/3)a+1)^2=9

Simplifying that got me 25a^2+27a-72=0.

This was the point where I knew I was wrong. Where did I go wrong/how do I fix it?

Can't you just use trig to find the 2 points? My sketch has two triangles with lots of known information about them...

 

[HallsofIvy]

Homework Statement

Find the points of tangency to a circle given by x^2+y^2=9 from point (12,9).

Homework Equations

dy/dx=-x/y
(what I've been able to come up so far)

The Attempt at a Solution

Taking the derivative I got dy/dx=-x/y

Let the unknown point of tangency be (a,b)

y-b=(-a/b)(x-a)

So this is to be the tangent line to the circle at (a, b)?

Simplifying that, I got:
by-ax=a^2+b^2

No, it's not. Multiplying both sides of y- b= (-a/b)(x- a) gives
b(y- b)= by- b^2= -ax+ a^2. Now add ax and b^2 to both sides to get
by+ ax= a^2+ b^2

a and b fall on the circle; the circle's equation is x^2+y^2=9; therefore, a^2+b^2=9

by-ax=9

by+ ax= 9

(12,9) is a point on this ^ line, so

9b-12a=9
b=(4/3)a+1

12b+ 9x= 9
b= (-4/3)ax+ 1

Substituting back into the original equation x^2+y^2=9,

a^2+((4/3)a+1)^2=9[/quote]
a^2+ ((-4/3)a+ 1)^2= 9

Simplifying that got me 25a^2+27a-72=0.

This was the point where I knew I was wrong. Where did I go wrong/how do I fix it?

 

[purplecows]

So I fixed that and even then, the answer was still way off. Can someone explain how to do this step by step with the final solution?

 

[SammyS]

So I fixed that and even then, the answer was still way off. Can someone explain how to do this step by step with the final solution?

What is the that to which you are referring?

To give "Step - by - step" is to give the solution.


Show what your steps are after fixing the that, so that we can help you.

 

[berkeman]

So I fixed that and even then, the answer was still way off. Can someone explain how to do this step by step with the final solution?

How about you post a sketch of the circle and two tangent lines? And label the two right triangles...

 

[Mentallic]

As berkeman has said, there will be two tangent lines that pass through an external point to a circle. This means you'll have two values of a, hence two values of b which gives you your points of contact on the circle $(a_1,b_1)$ and $(a_2,b_2)$.

And how would you expect the algebra to indicate to you that you have two solutions? Well, quadratics can have two solutions