[Locutus of Borg]

4 . Suppose that point P(a, b) is any point on the circle with center O(0, 0) and radius r. Suppose that
line l is perpendicular to segment OP at P. Prove that l is tangent to the circle as follows:
a. Show that the equation of l can be written as ax + by = r2.
b . Solve the equations of l and the circle simultaneously. Show that there is only one
solution.

I've already done a by saying that the slope of the radial line connecting the center and point P is b/a, therefore the slope of the tangent line must be -a/b
y - y1 = m(x - x1)
y - b = -a/b (x - a)
y - b = (-ax + a^2)/b
y = (-ax + a^2) + b^2) / b
by = -ax +a^2 + b^2
ax + by = a^2 + b^2
a^2 + b^2 = r^2 (Pythagorean theorem)
ax + by = r^2

but I'm lost on how to do b.

[FreeSpaceFreak]

"Show that there is only one solution" means that there's only one intersection between the tangent line and the circle: if there were two, they would intersect; if there were none, they wouldn't touch each other.

So, solve the system of two equations (line + circle) and two unknowns (x,y) and there should be only one possible outcome, the coordinates of the tangent point.

[Locutus of Borg]

Well yes. Every single one of my attempts has resulted in a dead end.

x^2 + y^2 = ax + by
x^2 - ax + y^2 - by = 0
x(x-a) + y(y-b) = 0

gets me nowhere

I've even solved for x and y individually

y = (r^2 - ax)/b AND (r^2 - x^2)^1/2

x = (r^2 - by)/a AND (r^2 - y^2)^1/2

[FreeSpaceFreak]

You're doing it wrong Eliminate x or y, not r.

[watsisname]

Use derivatives. 

[Locutus of Borg]

Unfortunately, I'm not allowed to use calculus to solve the problem.

[watsisname]

Ok, try figuring out the equation of the line.

Hint:  Use the point-slope formula.  You know the slope of the line already (it's tangent to the line from O to (a,b), and you know it must pass through the point (a,b).

You know the equation of the circle, so solve both equations simultaneously and you should find that the point (a,b) is the only solution.

[Locutus of Borg]

the equation of the line is ax + by = r^2

I had it in point slope before

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

I've substituted that in for (y-b) but I still don't have a solution

[Mongoose]

I'm bored enough to take a crack at this. Gimme a few minutes to take a look.

[Herra Tohtori]

Technically there's infinite number of parametric solutions because there's infinite points P(a,b) but for each P(a,b) there is only one solution for the tangent.

I don't know what you're supposed to prove in this problem because that's the definition of tangent for a circle.
For curves in general, definition of tangent is one local intersection, meaning the tangent only touches the curve once on a section that has same curvature. When curvature changes, as it does in odd-degree polynomial curves (3rd, 5th etc ), the tangent may intersect with the curve itself non-locally, but the curvature of a circle is constant so one intersection is the definition of the tangent.


If we have a circle

x² + y² = r²

and we substitute point P(a,b) on it as x=a and y=b, we get

a² + b² = r²

and the equation of the line being

ax + by = r²

we need to check if there are any other solutions than x=a and y=b for this equation.

So, the group of equations you need to look at is

a² + b² = r²
x² + y² = r²
ax + by = r²

...all of which are equal to square of the radius of the circle.

What you need to do here is prove that there are no other solutions than x=a and y=b for this group of equations...

[Locutus of Borg]

How should I go about that? Should I subtract two of those?

[Mika]

You have a set of equations:

(1) xa + by = r²

(2) x² + y² = r²

Solve x from (1) and substitute that to (2). Then solve y from the resulting equation. You should end up with a second degree polynomial that I assume you can solve.
The identity a² + b² = r² comes handy in there.

[Locutus of Borg]

(r^4 - 2r^2by + b^2y^2)/a^2 + y^2 = r^2

[Mongoose]

You have a set of equations:

(1) xa + by = r²

(2) x² + y² = r²

Solve x from (1) and substitute that to (2). Then solve y from the resulting equation. You should end up with a second degree polynomial that I assume you can solve.
The identity a² + b² = r² comes handy in there.

That's a hell of a lot neater than what I was trying.  I went with watsisname's suggestion and started with the point-slope form of the line instead of your (1), and I came up with this unsolvable algebraic mess. Granted, this method's producing a nice algebraic mess too, but at least it's far more comprehensible.

[Mika]

Seems about right to me, just keep on going. You know how to solve a quadratic equation? I would advise against dividing by a², though. It makes things easier later.

Just collect the terms under correct y's powers and write it down as Ay² + By + C = 0, solving it from the standard form is not too difficult.

[Herra Tohtori]

(r⁴ - 2r²by + b²y²)/a² + y² = r²

r² = a² + b²

substitute that into the equation...

[Locutus of Borg]

Seems about right to me, just keep on going. You know how to solve a quadratic equation? I would advise against dividing by a², though. It makes things easier later.

Just collect the terms under correct y's powers and write it down as Ay² + By + C = 0, solving it from the standard form is not too difficult.

The division by a^2 is part of the x identity

(r⁴ - 2r²by + b²y²)/a² + y² = r²

r² = a² + b²

substitute that into the equation...

(a⁴ + 2a²b² + b⁴ - 2a²by - 2b³y + b²y²)+ y² = a² + b²
--------------------------------------------------
a²

[Mongoose]

Yeah, I got a few steps beyond that before I gave up for the moment and went to replay some HL2.  It's great to know that my assignment work ethic is as fantastic as ever.

[Herra Tohtori]

(a⁴ + 2a²b² + b⁴ - 2a²by - 2b³y + b²y²)+ y² = a² + b²
--------------------------------------------------
a²

Well, there you have it.

Now simplify it into form

Ay² + By + C = 0

where A, B and C are the multipliers of y², y¹ and y⁰ for the second degree function. They should be a big mess of a's and b's and their exponents.

Then solve it normally with the quadratic formula:

y = -B ± Sqrt ( b² - 4 AC ) / 2A

...and you should get one (real) solution for the equation. Discriminant should end up as zero, and there should be your proof for there being only exactly one real solution, actually.



If you want to, you could include the complex solution and show that, in fact, there are more than one solutions for the equation group, but I guess the problem assumes solutions within real numbers. Nth degree polynomial functions always have N solutions, after all...

[Locutus of Borg]

????