Why the radius meets it at a right angle, argued from shortest distance

0:00Roll a wheel along a flat road, and look at the spoke that happens to be pointing at the ground. It looks square to the road. Every time. Whatever moment you catch it in. That is worth being suspicious of, because looking square and being square are two different things, and a picture cannot tell you which one you have got. Strip the wheel down to what it actually is. A circle. A road that touches it at exactly one point. A spoke running from the centre out to that point.

0:34So the claim hiding inside the picture is this. A radius drawn to the point where a tangent touches meets that tangent at a right angle. We are going to prove it. And we are going to prove it without measuring a single angle. Before anything else, separate what we are given from what we have to earn. We are given a circle, and a line that is a tangent to it. That word carries exactly one piece of information. The line and the circle share exactly one point.

1:11That is the whole hypothesis. Not that the line is horizontal. Not that it grazes gently. Not that it looks square to anything. One shared point. What we have to earn is the right angle. Notice how far apart those two things are. One is a statement about counting. The other is a statement about direction. Nothing inside a count knows anything about angles. Getting from one to the other is the entire job, and here is the whole of it.

1:45No other point of the line can be inside the circle. No other point can be on the circle. So every one of them is strictly outside, which makes the radius the shortest reach from the centre down to the line. And the shortest reach from a point to a line is the perpendicular one. Draw it. A circle with centre O. A line that touches it, and I will call the touching point P.

2:14Join O to P. That segment is a radius, so its length is exactly r, whatever r happens to be for this circle. Now pick any other point of the line at all. Call it Q. Q is a rival, a competitor to P for the title of nearest point of the line to the centre. Join O to Q as well. I am drawing that one dashed, because P's segment is the one we are defending and Q's is the one we are testing it against.

2:47And notice what I have not drawn. There is no little square at P. There will not be one for most of this video. That is deliberate. The right angle is the thing being proved. Draw it now and I have assumed the answer, and everything after it is theatre. First question. Could Q be inside the circle? Suppose it is. Q sits on the line, and Q is inside.

3:16Follow the line away from Q in one direction. It starts inside, and it cannot stay inside for ever, so somewhere it crosses the circle. Follow it the other way and exactly the same thing happens. That is two crossings. Two shared points. But a line sharing two points with a circle is a secant, and we were handed a tangent, which shares exactly one. So Q is not inside. Not because it does not look inside. Because if it were, the line we were given would have to be a different kind of line from the one we were given.

3:57Second question, and this is the one that gets skipped. Could Q be on the circle? Suppose it is. Then the line runs through P, which is on the circle, and through Q, which is also on the circle, and P and Q are different points. Two shared points again. A secant again. The same contradiction. It is worth saying why this step is not decoration. If we ruled out only the inside, we would know that Q is on the circle or outside it. And on the circle gives a length equal to r, not longer than r.

4:36The whole argument runs on a strict inequality. One or equal anywhere in the chain and the conclusion evaporates. Both exclusions are load bearing. Put the two together. Q is not inside, and Q is not on. So Q is strictly outside. And outside is not a vague word here. It has a length attached to it. A point is inside when its distance from the centre is less than r. It is on the circle when that distance is exactly r. It is outside when the distance is more than r.

5:17That is what those three words mean. So Q being outside says, in numbers, that O Q is greater than r. And O P is equal to r, because P is a point of the circle. Therefore O Q is greater than O P. That line is the hinge of the whole proof. The rival segment is longer, and not because it slants in the drawing. It is longer because Q is outside and P is on.

5:47Now the step that turns one fact into a theorem. Q was any point of the line other than P. I never said where. So that inequality holds for every one of them at once. Slide Q along the line and watch the length of O Q. It comes down as Q approaches P, it reaches its lowest value exactly at P, and it climbs again on the far side.

6:15Every rival reach is longer. So the radius O P is the shortest of all the segments from the centre down to that line. That is now a statement about a minimum, and a minimum is something you can argue with. We have turned a fact about counting into a fact about lengths, and we still have not measured an angle. One more ingredient, and it has nothing to do with circles.

6:43Take any point off any line. Drop segments from the point down to the line, as many as you like, spread all along it. Among all of them, one is shorter than the rest. The claim is that the shortest one is the perpendicular one. That is a separate result and it deserves to be treated as one. It is very easy to believe it is a definition. It is not.

7:10Distance is defined between two points. Speaking of the distance from a point to a line is a shortcut, and this result is exactly what licenses the shortcut. So let us prove it. It takes one line. Here is the point, here is the line, and here is the perpendicular dropped from the point onto it. Call its foot F. Now take any other point of the line. Call it R, and R is not F.

7:41Look at the triangle with corners at the point, at F, and at R. The angle at F is a right angle, because that is how F was built. So this is a right triangle, and the segment down to R is its hypotenuse, the side facing the right angle. The perpendicular is a leg. A leg of a right triangle is shorter than its hypotenuse. Always. So the perpendicular beats that segment.

8:12And R was any point of the line, so the perpendicular beats all of them, one at a time. It is the shortest, and it is the only shortest. You will often meet this argued the other way about. Assume the shortest one is not perpendicular, drop the true perpendicular anyway, and watch a leg beat the supposed champion. A contradiction, so the assumption was wrong. That works, and it is the same triangle doing the same work. But the result does not need contradiction. The direct comparison settles it in one line, and it tells you more, because it compares the perpendicular against everything rather than against one assumed winner.

9:03Now put the two halves side by side. From the circle we got this. The radius to the touching point is the shortest segment from the centre to the tangent line. From the triangle we got this. The shortest segment from a point to a line is the perpendicular one. The shortest one is unique, so those two segments are the same segment. The radius is perpendicular to the tangent.

9:31And only now do I draw the square at P. Look hard at what produced it. Not a protractor. Not the drawing. A count, one shared point, pushed through two exclusions into a strict inequality, then into a minimum, and then met by a fact about right triangles. That is what proving something means. The picture was true the whole time. It simply was not evidence. Two consequences follow almost free, and both get misquoted constantly.

10:06The first. At any point of a circle there is exactly one tangent. Be careful with exactly, because it is doing two jobs at once. That there is at most one is what we have just proved. A tangent there has to be perpendicular to the radius at that point, and there is only one line perpendicular to a given line at a given point. That there is at least one, that a tangent exists there at all, is a different question with a different answer. You get it by constructing one.

10:42From a point inside the circle there are no tangents at all, and from a point outside there are two. It is the points of the circle itself that get exactly one. The second consequence is a name. The line carrying that radius, continued on past the touching point, is called the normal to the circle there. The normal and the tangent are two different lines through the same point, square to each other. The normal runs through the centre. The tangent never does.

11:18Now use it, because this theorem earns most of its keep as a piece of arithmetic. A circle of radius five. A point Q sitting out on a line through the centre, twelve away from the centre. A tangent drawn from Q touching the circle at P. How long is the tangent, from Q to P? The theorem hands you a right angle at P, so the triangle O P Q is right angled there.

11:48Which side is the hypotenuse? The one facing the right angle. The right angle is at P, so the hypotenuse is O Q. The twelve. The twelve is the big one, so we subtract. P Q squared is a hundred and forty four take away twenty five, which is a hundred and nineteen. So P Q is the square root of a hundred and nineteen, which is a little under eleven and is not a whole number.

12:18Now the trap, and it is the commonest error anywhere near this theorem. Add instead of subtracting and you get a hundred and forty four plus twenty five, which is a hundred and sixty nine, and a beautifully tidy thirteen. Thirteen is a genuine answer to a genuine question. It is the answer when the twelve is a leg and the distance to the centre is what you are looking for.

12:46It is not the answer to this one. Deciding which length faces the right angle is the whole of the arithmetic here, and a tidy answer is not evidence that you decided correctly. So here is the argument, end to end, on one board. A tangent shares exactly one point with the circle. Any other point of that line cannot be inside, because then the line would cross twice. It cannot be on the circle, because then the line would share two points.

13:21So it is strictly outside, so it is farther from the centre than the radius is. Every rival is farther, so the radius is the shortest reach from the centre to the line. The shortest reach from a point to a line is the perpendicular one. So the radius is perpendicular to the tangent. No measurement anywhere in that. No protractor, and no appeal to how the drawing looks. The spoke on the road was square all along. The difference is that now you know it, instead of seeing it.