Lines and circles tend to avoid each other, because they kind of freak each other out. Lines don't care for the weird curviness a circle has, while circles are mystified by the fact that a line goes on forever and ever and ever...
Anyway, circles and lines don't get along too well, but since they're at the mercy of whoever draws them (to some extent), they do have to interact sometimes.
Throw a line down onto a plane with a circle in it and one of three things will happen. If they're lucky, the line and the circle won't intersect at all. They'll each go about their business, perhaps warily eyeing each other, but not sharing any points.
You might end up with the line intersecting the circle at two points.
Or, once in a blue moon, you might see the line just "touching" the circle, intersecting it at exactly one point.
We have special names for these last two cases. If a line intersects a circle at two points, then the line is a secant of the circle. If a line intersects a circle at exactly one point, then the line is tangent to the circle. (You can think of a tangent line as just barely touching the circle.)
How many different tangents can there be at one given point on a circle? That's not an easy question to answer, is it? We feel like there ought to be only one tangent line for each point on the circle, but at the same time we don't really know why that should be the case.
How about a postulate? That sounds good. We'll just assume that for any point on a circle, there's only one line tangent to the circle at that point. In math, assumptions are always true, unless they contradict statements already known to be true. So let's just make this assumption and hopefully nothing bad will happen.
Given ⊙O and point P on ⊙O, there is exactly one line tangent to ⊙O at P. Remember, "exactly one" means, "no less and no more than one."
By the way, the phrase "going off on a tangent," comes from geometry. The idea is that when you're on-topic, you're following the curve of the circle, but at some "point," you veer off into space along some line and have trouble getting back. Sometimes, to the point of getting lost in space.
How's this for a tangent? It's a guy with a trenchcoat and a cowboy hat using a sling to swing a tennis ball over his head in slow motion. The only way it could be more tangential is if he releases it.
Oh wait. He does.
When he releases it, the ball travels along a straight line, tangent to the circular path. Here's an overhead diagram of this situation. The string is a radius of the circular path of the ball.
Notice that when the sling man wants to hit a target directly in front of him, he releases the ball when the string is pointing directly off to the side. If he wanted to hit a target over to his left, he would release the ball when the string was straight out in front of him.
In other words, when the ball is released, it travels along a path perpendicular to the string's position at that moment. In fact, a line perpendicular to the radius at a point on the circle is always a tangent line. We feel a theorem coming on.
Given a circle ⊙O and a point P on ⊙O, if a line m through P is perpendicular to the radius OP, then m is tangent to ⊙O at P. Let's start by drawing a picture of the situation, adding in a point Q on m somewhere other than P.
Wherever Q is on m, the distance OQ must be greater than OP, because the shortest path between a point and a line is a perpendicular segment. By definition of the exterior of a circle, Q (and therefore any point on m other than P) must be in the exterior of ⊙O. Therefore, we know for sure that m intersects ⊙O at no point other than P (where it does intersect ⊙O). Therefore, by definition of a tangent line, m is tangent to ⊙O.
We can use this theorem to prove its converse. Sneaky, but totally legal.
If a line m is tangent to ⊙O at P, then m is perpendicular to OP. Let's start by remembering that for any line and a given point on that line, there is exactly one line through that point that is perpendicular to that line. Using that principle, let's draw the unique line q that is perpendicular to OP at point P. By the theorem we just proved, q is tangent to ⊙O at P.
But wait a second. Now we have two lines, m and q, that are both tangent to ⊙O at P. This violates the Postulate that says there can only be one line tangent to a given circle at a given point…unless m and q are the same line! In other words, m = q. Since m = q and q is, by our definition, perpendicular to OP, m must also be perpendicular to OP.
We can combine these two into one biconditional theorem. Let's call it the Perpendicular Tangent Theorem. Or, you know, Hubert. Your call.
Given a line m through a point P on ⊙O, m is perpendicular to OP if and only if m is tangent to ⊙O at P. In other words, a tangent line is always perpendicular to the circle's radius at the point of intersection.
Now you know all you need to know in order to be an expert sling-hunter, just like that trenchcoat-wearing cowboy. Shmoop is not responsible for any injuries you may sustain or inflict while practicing.
Lines m and n are tangent to ⊙O at points P and Q, respectively. Where must P and Q be located on ⊙O so that lines m and n are parallel to each other?
According to the Perpendicular Tangent Theorem, tangent lines are always perpendicular to a circle's radius at the point of intersection. In other words, m is perpendicular to OP and n is perpendicular to OQ. In order for m and n to be parallel (never intersect), this means ∠POQ has to be 180°. In other words, P and Q have to be at opposite ends of ⊙O (meaning that PQ is a diameter of ⊙O).
Let's raise the stakes a little bit. So far, we've been dealing with one circle at a time. Suppose we have two circles. Whoa.
Can we draw a line that is tangent to both these circles? Sure we can. As a matter of fact, we can draw four such lines, which we call common tangent lines, or "common tangents" for short.
Not all pairs of circles have four common tangents. Take a look at these two circles. They're tangent to each other, meaning they share exactly one point in common. They only have three common tangent lines. See?
What is an example of a pair of circles that has zero common tangents?
Zero? As in, no common tangents…at all? None? Is that even possible?
Actually, yes. See? Tangents to the outer circle won't touch the inner circle at all, and tangents to the inner circle will always be secants of the outer one. No matter where we draw any tangent, we'll never find a common one. How depressing.
A cross-section of an idealized ice-cream cone might look like this.
Here, the scoop of ice cream is a perfect sphere, and the cone is, well, a perfect cone. (We'll talk more about spheres and cones later on.) This is the ISO 440 standard model for ice cream cone design, used by frozen dessert engineers all over the world. (Disclaimer: this is completely and utterly false.)
In this cross-section, the ice cream is a circle and the sides of the cone are line segments, each of which intersects the circle at exactly one point.
If a line segment is a segment of a tangent line and has one of its endpoints on ⊙O, then the line segment is tangent to ⊙O. We sometimes call it a tangent segment.
Notice that in the ice cream diagram, the two segments that make up sides of the cone share an endpoint at the bottom of the cone. This gives those segments a special property.
Given a point P in the exterior of a ⊙O, if segments PQ and PR are tangent to ⊙O at points Q and R respectively, then PQ is congruent to PR. In other words, any two segments that share an endpoint and are tangent to the same circle are congruent. We'll call this the Endpoint-Tangent Theorem.
How can we prove it? We'll call on our old buddies, congruent triangles, to help us out. Not just any congruent triangles, though: congruent right triangles. But first we need some right triangles. Let's start by drawing segments OQ, OR, and OP.
Since QP is a tangent, it's perpendicular to OQ, and segments OR and RP are perpendicular as well. This makes ∠OQP and ∠ORP both right angles. We now have two right triangles with a common hypotenuse, OP. We know that OQ is congruent to OR because they're both radii of the same circle. By the Hypotenuse-Leg Theorem, ΔOQP is congruent to ΔORP. Since congruent triangles have corresponding congruent parts, PQ is congruent to PR.
It's the Endpoint-Tangent Theorem that ensures that ice cream stays in its cone, where it's supposed to be. Gravity helps, too.