Finding Points of Tangency for the Unit Circle

[Cascadian]

Hi I'm trying to study over break, this isn't an exact quote but its the part of the problem I need help with. Thanks.

Homework Statement

Draw the unit circle and plot the point P=(3,2). Observe there are TWO lines tangent to the circle passing through the point P. Lines L1 and L2 are tangent to the circle at what points?

Homework Equations

The Attempt at a Solution

I tried plugging in the point into a point-slope formula, it was kind of a dead end for me.

 

[phinds]

draw a figure and show what you know

 

[haruspex]

Pick a point on the unit circle, (cos α, sin α) say. Write down the equation of the tangent there.

 

[phinds]

Pick a point on the unit circle, (cos α, sin α) say. Write down the equation of the tangent there.

How about if we get him to draw a diagram first and then help him from there?

 

[Mandelbroth]

Hi I'm trying to study over break, this isn't an exact quote but its the part of the problem I need help with. Thanks.

Homework Statement

Draw the unit circle and plot the point P=(3,2). Observe there are TWO lines tangent to the circle passing through the point P. Lines L1 and L2 are tangent to the circle at what points?

Homework Equations

The Attempt at a Solution

I tried plugging in the point into a point-slope formula, it was kind of a dead end for me.

Equation for unit circle is y^2 + x^2 = 1. Equation for line passing through point P=(3,2) is y - 2 = m(x-3).

These facts might be helpful. They might also fit under that bold region labeled "2. Homework Equations ".

So, the question is this: if you want to find 2 values for x and y such that both equations are satisfied, what can you do with the two equations?

 

[Cascadian]

Hi everyone, thank you for the input. Let me clarify my situation, I uploaded a diagram; the approach I had been working on was the one Mandelbroth had mentioned. I had y = mx-m3+2 and y^2 + x^2 = 1, when I plugged the first equation into the second I wasn't sure what to do especially because there were now 3 unknown variables, y, x, and m. That's where I'm at now.

 

[haruspex]

Hi everyone, thank you for the input. Let me clarify my situation, I uploaded a diagram; the approach I had been working on was the one Mandelbroth had mentioned. I had y = mx-m3+2 and y^2 + x^2 = 1, when I plugged the first equation into the second I wasn't sure what to do especially because there were now 3 unknown variables, y, x, and m. That's where I'm at now.

The problem with htat approach is that nowhere have you specified that the line is supposed to be a tangent. Do you know how to write down the generic equation for a tangent to circle at a given point, (cos θ, sin θ) say?

 

[Mentallic]

Hi everyone, thank you for the input. Let me clarify my situation, I uploaded a diagram; the approach I had been working on was the one Mandelbroth had mentioned. I had y = mx-m3+2 and y^2 + x^2 = 1, when I plugged the first equation into the second I wasn't sure what to do especially because there were now 3 unknown variables, y, x, and m. That's where I'm at now.

Yes, so if you plug y=mx-3m+2 into x^2+y^2=1 then you'll have an equation in terms of x and m. It will be a quadratic in x, so if you solve the quadratic for x, you'll have a solution that is in terms of m.

Now, what you need to do is to understand what the solution to the quadratic is telling you. What does the discriminant in the quadratic formula say?

The problem with htat approach is that nowhere have you specified that the line is supposed to be a tangent. Do you know how to write down the generic equation for a tangent to circle at a given point, (cos θ, sin θ) say?

There is no problem with that approach. He'll be picking the values of m that are tangents to the circle soon enough.

 

[symbolipoint]

Distance from O to point (3,2) uses distance formula.
Distance from any point on circle to O is 1 for a unit circle.
Two points on the circle are tangency points, upper may be called B, (x₂, y₂).
Angle of triangle at B is right-angle, so POB is right triangle.
Distance from B to point (3,2) uses Pythagorean Theorem, because you know the other two triangle side lengths.

Use DISTANCE FORMULA again for B to point (3,2).

Two equations and unknowns:
x²+y²=1
BP=sqrt((3-x)²+(2-y)²)