Fixed distance from a fixed point, turned into an equation

0:00Four curves come out of a cone, and each of them gets a definition. The parabola's definition names a point and a line. The ellipse's and the hyperbola's each name two points. The circle's names one thing: a point. A circle is every place at one fixed distance from one fixed point. One point and one number, and that is the whole of it. Which is why, out of the four, this is the one where there is nothing to derive.

0:32The definition already tells you the distance. The distance formula already tells you what a distance is. Putting those two side by side takes a single line. Everything worth learning in this topic is the trip back the other way. Start with the definition as a picture. Mark a point, and call it the centre. Now mark every place that sits a fixed distance away from it. Three of them, thirty of them, all of them - the lengths from the centre are all the same length, and that sameness is the only thing being asked.

1:11Notice what is not in that description. No axes, no coordinates, no equation. A circle is a set of places, picked out by comparing one distance with another. That distinction is going to matter, because an equation is a different kind of object altogether. So every equation in this video gets scored against a set of places built from the definition alone - which is what lets an equation be wrong.

1:43Now put the picture on axes. The centre is somewhere. Call its two coordinates h and k. A place on the circle is somewhere. Call its two coordinates x and y. The definition says the distance between those two is r. The distance formula says what that distance is: drop a right angle between them, and the two legs are x minus h, and y minus k. So the square of the distance is x minus h, squared, plus y minus k, squared.

2:18The definition says that distance is r. The formula says what it is. Put the two together and you have the square root of that sum, equal to r. That is already the equation. What is left is tidying. Square both sides, and the root disappears. Squaring is a step worth being suspicious of, because it can let things in. Solve a squared equation and you can end up with answers the unsquared one never had.

2:50Here it lets nothing in, and there is a reason. A distance is never negative, and neither is r. Square two numbers that are both not negative, and you can always get back. That was measured rather than trusted. Nearly one hundred and ninety thousand places were tested against both statements at once - the one with the root, and the one without - and the number the squaring let in is none.

3:19The same test, run on a statement whose right-hand side is allowed to go negative, let in one thousand two hundred and eight. So squaring is free here. It is not free in general, and the difference is a sign. What survives is the template. x minus h, squared, plus y minus k, squared, equals r squared. Put the centre at the origin and both letters go to nothing, which leaves x squared plus y squared equals r squared.

3:54Now the part that catches everyone. The template SUBTRACTS the centre's coordinates. So a centre with a negative coordinate produces a plus sign inside the bracket. Centre at minus three, two, with radius four, gives x plus three, squared, plus y minus two, squared, equals sixteen. Of eighty-one centres tried here, seventy-two have a coordinate that turns a bracket's sign around. And in eighty of the eighty-one, writing the centre's own sign straight into the bracket gives a different equation from the right one.

4:33This is not a rare trap. It is nearly all of them. Now the check that keeps all of this honest. One thousand two hundred and ninety-six circles were built as sets of places, straight from the definition, and each was handed the equation its own derivation produces. Then every place in a window was asked two separate questions: is it in the set, and does the equation come out at nothing there.

5:04Three hundred and seventy-four thousand five hundred and forty-four askings, and the number of disagreements is none. That would prove less than it sounds if the scoring could not catch a wrong equation. So every circle was handed a decoy as well - the same equation, one unit out. Of the one thousand two hundred and ninety-six decoys, four hundred and sixty belong to circles the window can actually see, and all four hundred and sixty were caught.

5:37The other eight hundred and thirty-six are circles the window never touches, where a decoy one unit out cannot be told from the real thing. That gets reported, rather than counted as a pass. Now the trip back. You are handed x squared plus y squared, plus eight x, plus ten y, minus eight, equal to nothing - and asked for the centre and the radius. Nothing in that looks like the template. The brackets have been multiplied out.

6:12So put them back. Take the x terms: x squared plus eight x. To be a bracket squared, that wants sixteen more. Take the y terms: y squared plus ten y. That wants twenty-five more. Move both across to join the eight that was already there. Eight plus sixteen plus twenty-five is forty-nine. Centre at minus four, minus five. Radius seven. Why did that work? Because it is not a technique at all. It is the derivation run backwards.

6:49Multiplying out x minus h, squared, is what put those terms there in the first place: it leaves minus two h times x, and an h squared behind. So the number in front of x is always minus twice the centre's first coordinate. Halve it, turn the sign round, and the centre falls out. The routine that does this here was never told what it was looking for. It gets six numbers and nothing else.

7:20Five thousand one hundred and eighty-four equations were put through it - including ones multiplied through by two, by minus three, and by a half beforehand - and the number where it failed to recover the centre and the radius is none. Run it the other way as well, taking an equation apart and rebuilding it, and of one thousand two hundred and ninety-six, the number that came back different is none.

7:50It does not always succeed, and where it refuses is exactly what the usual account leaves out. It complains three ways. If the two squared terms carry different sizes, there is no circle to find - the shape is something else entirely. If there is an x-times-y term, the same. And if there is no squared term at all, there is nothing to complete. Ten equations were offered to it by hand. It read a centre out of four of them, and made three distinct complaints about the rest.

8:26The first complaint has a cure worth knowing. If both squared terms carry a two, divide the whole equation by two before you start. Two x squared plus two y squared minus x, divided through, gives a centre at a quarter and nothing, and a radius of a quarter. There is a second condition, and it is about the number left at the end. The completion hands back something that is supposed to be the radius squared. It does not have to be positive.

9:01Seven hundred and two equations of exactly circle shape were built and asked what they actually draw. Three hundred and twenty-four draw a circle. Fifty-four draw a single point, because the radius is nothing, and the only place at no distance from the centre is the centre itself. And three hundred and twenty-four draw nothing whatever, because no place is a negative distance from anything. Three kinds, and on every one of the seven hundred and two the definition and the equation agreed about which.

9:40It happens with a radius written as an expression too. Take the square root of a squared minus b squared: over eighty-one pairs of whole numbers, thirty-two give a circle, seventeen give a point, and thirty-two give nothing. Now the other direction again. You are given conditions, and asked for the circle. Count before you compute. The template has three unknowns: two for the centre, one for the radius. So you need three independent conditions, and no fewer.

10:16Two points on the circle, plus a line the centre has to lie on, is three. Here is that counted rather than solved. Over a grid of centres, the number lying the same distance from both given points is twenty - a whole line of them, which is two conditions doing what two conditions do. Add the third, that the centre sits on a given line, and twenty drops to one: seven tenths, and thirteen tenths.

10:46And notice that the radius squared was never solved for. Subtracting one point condition from the other removes it. It falls out at the end, on its own. Three conditions do not always leave one answer. Radius five, centre somewhere on the flat axis, passing through the place two, three. Search the same grid and two centres come back rather than one: minus two and nothing, and six and nothing.

11:17Two different circles, both of radius five, both through that place, both centred on that axis. A question that asks for THE circle here is asking for something that is not there. One more worth having. A circle through the origin that crosses the axes at four and at six. Three places, so one centre - two, three - and the equation comes out with no constant term at all, which is exactly what passing through the origin means.

11:50Of six hundred and eighty-six triples of places offered a circle through all three, eighty-four were refused, because those three lie on one line and no circle passes through them. Last, the cheapest question in the whole topic. Is a given place inside the circle, outside it, or on it? You do not need a drawing. You need one substitution. Work out the squared distance from the centre, and compare it with the radius squared. Smaller is inside, equal is on, bigger is outside.

12:28Five hundred and seventy-eight places were tested that way. One hundred and thirty-eight came out inside, twenty-four came out on, and four hundred and sixteen came out outside - with none at all that came out as two of the three, or as none of them. Minus two and a half, three and a half, against a radius squared of twenty-five: the squared distance is thirty-seven halves, which is under twenty-five. Inside.

12:59And that is the topic. One definition, one formula, and one line of algebra. The only real work left is learning to read that line backwards.