Geometry of the Line (4): Circles

Time to go back to straight line geometry for a bit.

With knowledge of how to calculate the gradient, the perpendicular gradient, the mid-point and the length of the line, you've also done quite a lot of co-ordinate geometry of the circle. This is an interesting concept. A circle on a set of cartesian co-ordinates.

You start from the premise of a circle with centre (0,0). The equation of this circle is x² + y² = r². If you study a diagram of it, or even draw one yourself, the reason for this becomes clear. Like a lot of co-ordinate geometry, it's based on Pythagoras. In this case, drawing a triangle inside the circle to a point somewhere on the circumference. Say that this circle with centre (0,0) passes through the point (3,4).

The line from the centre to the point functions as the hypoteneuse of the right angled triangle, while line x measures the distance travelled horizontally to the point, and the line y measures the vertical distance to the point.

The equation of this particular circle then tells us the radius.

3² + 4² = r²

or 9 + 16 = r²

= 25

r = 5.

The radius of the circle is 5.

Things get a little more complicated if the circle's centre is not (0,0). Say we have a circle with centre (2,3). We still devise the equation of the circle based on its distance from (0,0).

This circle's centre has moved 2 units x-wards and 3 y-wards (across and up) from the origin. Any calculation of the radius needs to take into account the fact that we are already some way from the origin.

So if the circle passes through the point (5,8), the distance from the centre to the point will be 5-2 and 8-3 (otherwise we're talking about including space outside the circle).

So the equation will be: (x-2)² + (x-3)² = r².

In this case: 9 + 25 = 34

So r = √34

It is perfectly acceptable to leave the answer as a surd at C1, indeed often you have to.

Usually you are given the equation, say (x-4)² + (y-7)² = 4

or something.

The centre of the circle is ALWAYS THE OPPOSITE OF THE NUMBERS IN BRACKETS, in this case (4,7). Worth remembering. If it is (x-a)² + (y-b)² = r² the centre is always (a,b).

Gradients of Perpendicular Lines

If a line intersects another one such that the two are perpendicular,and the gradient of the other is known, it is a piece of cake to work out the unknown gradient.

The gradients of perpendicular lines have a product of -1.

Our line from the previous post has gradient 12-6/8-3 = 6/5 Its perpendicular bisector will therefore have gradient of (using the common letter m to stand for unknown gradient):

m₁xm2=-1

6/5xm₂ = -1

m₂= -1 ÷ 6/5

At this point use the fraction division rule:

m₂= -1 x 5/6

= -5/6

The gradient of the perpendicular bisector is -5/6.

The really cool bit is this. If you have two co-ordinates of the first line, say in the post below (3,6) and (8,12); then you can now easily find the equation of the perpendicular bisector.

If it bisects the first line it must pass through the mid-point: x1+x2/2, y1+y2/2

or 3+8/2, 6+12/2

which is: (5 1/2, 9)

Plug what you know of this line into y - y₁ = m(x - x₁).

Or: y - 9 = -5/6(x - 11/2)

I've turned 5 1/2 into 11/2 for ease of calculation.

y -9 = -1/3x + 55/12

y = -1/3x + 163/12

This is a bit messy so multiply through by 12:

12y = -4x + 163



Lovely.

NOTE: I updated this post because it was wrong: it is still a bit more tricky than most of the questions you will get at C1. Most of these type questions resolve into a line with a constant of no more than about 20. But theoretically one like this could come up.

Geometry of the Line (3): Length

It is a fairly simple job to calculate the length of a line. So far I've shown the mid-points, gradients, and equations. This is similar stuff, and based on the principle of Pythagoras, as for gradients.

Take a line with two points (3,6) and (8,12) lying on the line.

We can see any straight line on a graph as being equivalent to a hypotenuse of a right angled triangle.

Now Pythagoras says that the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Looking at the diagram:




The line we want to measure is the hypotenuse and the other two sides are clearly horizontal and vertical. So the length of the horizontal line is easy - just x₂ - x₁ and the length of the vertical line the same but with ys. y₂-y₁

This is where Pythagoras comes in. We square the lengths of the two lines and add them together. This gives us the square of the hypotenuse.

We write it like this:

√(x₂-x₁)² + (y₂-y₁)²

In this case:

√ (8-3)² + (12-6)²

√ 25 +36

√ 61

This will be a surd so we can leave the answer as √61 for the purposes of C1.

Geometry of the Line (2) The Equation of the Line

The equation of the (straight) line isn't difficult. In fact it's dead easy.

The reason is this. Any curve or line has a statement that describes what happens to a y-co-ordinate when you plug in a given x-value. This is why we talk of f(x), because the y co-ordinate is basically the function of the x under certain conditions, which differ for various lines and curves. With this statement you have the potential to describe the co-ordinates of the curve or line completely knowing only the x-value.

The equation is simply the statement of what happens to the y co-ordinate when you select any x co-ordinate you like. It contains the gradient of the line within it, because clearly where the y-value is will depend on how steep the line is.

We often hear about y=mx + c , where m=gradient, and c= y-intercept (or where the line crosses the y-axis).

It is fine and dandy, to be sure.

But for MPC1 purposes the following equation is also useful, not least because solving it leads to y=mx+c.

(y - y₁) = m(x - x₁)

Where (x,y) is any, unknown point on the line and (x₁, y₁) is a known point.



For example: take a line passing through (2,4) and (6,10).

Firstly, work out m. 6/4. or 3/2.

Plug that into the equation: (y-4) = 3/2(x-6) or y-4 = 3/2x - 9 or y= 3/2x -5.

Suddenly you have your y-intercept (-5) and your y =mx+c.

The only thing is, you need the gradient, so you need to know two points on the line from the start.

Geometry of the Line (1) The MidPoint of the Line

Well I'm not sure that I'll bother to start with a definition of a line, although Barry Mazur, in his book that tied up my guts, Imagining Numbers, quotes Euclid: "one and only one line passes through any two distinct points on the plane". That makes sense to me.

For the purposes of this post the line will be assumed to be straight, and two dimensional. We are not going to muck about with curves and stuff.

So.

We have already seen that if you know two co-ordinates on the line you can calculate its gradient. This is very difficult to do with only one set of co-ordinates, unless you have a perpendicular line handy (which we don't, at the moment - that might come later).




If your line begins at 2,2 and ends at 8,4, as above, you need to think "How far has it gone x-wise and y-wise?" Well it has gone 6 units in the x direction and 2 in the y direction.

To find the midpoint of this line you can either just say well mid-way through a movement of 6 is 3, and mid-way through a movement of 2 is 1. So 2+3 = 5 (the x co-ordinate of your mid-point) and 2+1=3 (the y co-ordinate). Your midpoint is (5,3).

This isn't practical for a lot of lines which are (3 1/2, 5 1/2) and so on. So instead we use a simple formula:

The midpoint is, (x₁ + x₂)/2, (y₁+y₂)/2

So: (2+8)/2, (2+4)/2

Thus: 5, 3

The midpoint is (5,3), as above.


The midpoint is useful for all sorts of knotty problems involving perpendicular bisectors and stuff.

Geometry of the Line

Writing yesterday's post about gradient has given me an idea for a series of posts, which really ought to have predated the dy/dx one, on the C1 basics of geometry of the line of which gradient is a part. So, although I won't do it tonight, look out for a couple of posts on: equation of a line, finding the midpoints, gradients of perpendicular lines and so on.

Yes I know it isn't that exciting but you know, it is quite interesting really.