Circle, tangent line, and a point not on the circle

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Mar 30th 2011, 12:31 PM
iflyboats

Circle, tangent line, and a point not on the circle

http://www.webassign.net/userimages/...gentlines1.jpg

Suppose you have a unit circle and a point P with coordinates (4, 2). The task is to find the tangent points.

Here are my steps:

1. Name the coordinates of the tangent point in quadrant II as (r, s).
2. The slope of the tangent line is m = (s/r)
3. Rewrite the tangent point as (r, sqrt(1-r^2))

This is as far as I get before everything breaks down. Am I right so far?

4. My next step is to try to solve for r by using y = mx + b and the two points. But I end up with an ugly equation involving rational terms, radial terms, etc. and always end up with the wrong answer even when I cheat with a graphing calculator. Any idea what I'm doing wrong?
Mar 30th 2011, 12:54 PM
Plato

I would approach it differently.
Suppose that $(p,q)$ is a point of tangentacy.
The slope of a tangent at that point is $\dfrac{-p}{q}$ .

The slope using the point $(2,4)$ the slope is $\dfrac{q-2}{p-4}$ .
We know that $p^2+q^2=1$ . You can solve that system.
Mar 31st 2011, 09:09 AM
mathfun

A very simple solution. Let M(x,y) the tangency point.
Then

$\overrightarrow {OM} \cdot\overrightarrow {MP} = 0 \Rightarrow \left( {x,y} \right)\cdot\left( {4 - x,2 - y} \right) = 0 \Rightarrow 4x - {x^2} + 2y - {y^2} = 0$
$\Rightarrow 4x + 2y = 1$ which is the polar line.

The tangency points are the solutions of the system $4x + 2y = 1$ and ${x^2} + {y^2} = 1$
Mar 31st 2011, 09:09 AM
iflyboats

Still not having any success.

First, since the point of tangency is in quadrant 2, shouldn't the slope of the tangent line be positive? It seems to me that the slope of a perpendicular line from the origin would be $(-q/p)$ , and the slope of the tangent line would be $(p/q)$ .

So I obtain two equations for the slope of the tangent line, $(p/q)$ and $(2-q/4-p)$ .

Next I use the circle equation to solve for one of the coordinates in terms of the other: $q = sqrt(1 - x^2)$ . I plug this in to the slope equations above and solve for p. My answer is p = -0.63589, q = 0.77178. Unfortunately the answer is not correct. Any ideas what I'm doing wrong?
Mar 31st 2011, 11:31 AM
Plato

I don't see your error, but I get $p \approx - 0.236$ .
Mar 31st 2011, 01:40 PM
iflyboats

I see what I did wrong. Thanks guys.