How to find the slope of a line given one point and the fact that it runs tangent to a circle

Question

There's a circle with radius = 8 feet

The center of the circle is the origin

A person is situated at (12,0) and their line of sight runs tangent to the circle

The problem is to figure out which portion of the y-axis is obstructed by the circle for the person situated at (12,0)

The hint they give me is: Let $a$ be the x-coordinate of the point where line of sight is tangent to the circle; compute the slope of the line using two points (the tangent point and (12,0). On the other hand, compute the slope of line of sight by noting it is perpendicular to a radial line through the tangency point. Set these two calculations of the slope equal and solve for $a$.

To me that makes the tangency point $(a,y)$ as the hint want's me to substitute $a$ for $x$ and I also don't know $y$. Ok so then compute the slope given this scanty info I guess. I get:

$$\frac{-y}{12 - a}$$

Then the hint wants me to set this equal to it's negative reciprocal as that would be the slope of a line perpendicular to this one, so:

$$\frac{12 - a}{y}$$

And then setting them equal almost but not even close actually gives me a quadratic equation:

$$-y^2 = a^2 - 24a + 144$$

So I don't know what to do. Any hints about what I 'm doing wrong would be great, any hints about how to properly interpret the hint/problem is appreciated. Also, I realize there would be two lines of sight that are tangent (one in a upper quadrant and the other in a lower) but just focus on the one in the upper or whatever makes it easier for you.

Answer 1

See the sketch. We know $OT = 8, OP = 12$. So can you find $PT$ using Pythagoras?

Now using the slope of line through $AP$, you should notice that

$\displaystyle \frac{OT}{PT} = \frac{AO}{OP}~$

You know $OT, PT$ and $OP$. So can you find $AO$? Now the same length is obstructed below x-axis.

Answer 2

The slope of the tangent line is, as you note, $m = \frac{-y}{12-a}$.

The slope of the radius can be calculated directly because you have two points - the centre at $(0, 0)$ and the point of contact at $(a, y)$. Notice that the value of 12 doesn't seem to appear at all, so you should check that one carefully.

When you equate those two, you will get something that is quadratic in both $a$ and $y$. This is obviously something you want to get away from as quickly as possible, and that's where I'd suggest thinking about what other information you have about the point $(a, y)$. For example, is there an obvious equation that relates $a$ and $y$, maybe with some quadratic terms that would cancel out the ones you get from the gradients?

Answer 3

The radius is perpendicular to the tangent. Which means that the slopes of the lines are negative reciprocals of each other.

$-\frac {a}{y} = \frac {y}{a - 12}$

$a^2-12a = -y^2\\ a^2 + y^2 = 12a$

The next part is that the point $(a,y)$ on the circle. And the radius of the cirlce is 8.

$a^2 + y^2 = 8^2$

Now we can substitute between the two equations.

$64 = 12 a\\ a = \frac {16}{3}\\ y = \sqrt {64 - a^2} = \frac {\sqrt {320}}{3} = \frac {8\sqrt{5}}{3}$

$\frac {y}{a-12} = \frac {\frac {8\sqrt {5}}{3}}{ \frac {16}{3}-12} = \frac {8\sqrt 5}{16-36} = -\frac {2\sqrt 5}{5}$