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

1. Mar 29, 2014

### purplecows

1. The problem statement, all variables and given/known data
Find the points of tangency to a circle given by x^2+y^2=9 from point (12,9).

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

3. 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?

2. Mar 29, 2014

### Staff: Mentor

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

3. Mar 29, 2014

### HallsofIvy

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

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

by+ ax= 9

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

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

4. Mar 29, 2014

### 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?

5. Mar 29, 2014

### SammyS

Staff Emeritus
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.

6. Mar 30, 2014

### Staff: Mentor

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

7. Mar 30, 2014

### 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