Given the currency of the film, it's relevant to look at some examples from analytic geometry and offer up some problems too, What is analytic geometry? It is basically that area of mathematics that applies algebra and calculus to geometry in order to work out the locus of points for various figures, always expressible in the form F (x, y) = 0.
Once we generate the curve F(x,y) = 0 we can then study its peculiar geometric properties (for example those for a parabola differ from those for an ellipse) as well as find the equation of a curve once its geometric properties are known. All of these things we will do and I will challenge readers to do in assorted problems!
Probably the most straightforward equation and curve to begin examining is the circle. Let us consider one defined by the equation:
x 2 + y 2 = 4
This is very simple to plot as a graph and I invite readers to try. You will end up with the figure shown below:
This is actually an equation of basic form: x 2 + y2 = r2
where r denotes the radius. Since the square root of 4 (= r2 ) is 2, then we see from the graph the radius is 2.
But this simple example is deceiving because it may give the impression that all equations of the circle consist of three terms, and all circles are centered at the origin (0, 0). In fact, most interesting cases are circles which are off -center, and also of more complicated equations.
Consider, for example, the equation:
x 2 + y2 + 2x - 4y = 11
Is this a circle? Yes, it is, but we must work to get it into the correct form to analyze, or graph.
First, write the left side as:
x 2 + 2x + y2 - 4y
Now complete the square for the expression in x and y to get:
(x + 1) 2 + (y - 2) 2 = x 2 + 2x + 1 + y2 - 4y + 4
Re-configure the entire equation paying attention to terms added to both sides:
(x 2 + 2x + 1) + (y2 - 4y + 4) = 11 + 4 + 1 = 16
And if we now write in the most general form using coordinates (h, k) for the center:
(x - h)2 + (y - k ) 2 = r 2
We see at a glance, that h = -1 and k =2 with the radius r = the square root of 16 or 4. We can check this by doing the graph:
Inspection of the above graph shows that indeed (h, k) = (-1, 2), the coordinates of the center and the radius is 4 units.
There is also another, longer way to approach the problem using a generic analytical expression, e.g.
Ax 2 + Ay2 + Dx + Ey + F = 0 (A not equal 0)
This can be reduced to the earlier equation (x - h)2 + (y - k ) 2 = r 2
by completing the square, which is left as an exercise for the energetic reader.
When you do this you should obtain the general, analytical expression for the radius:
r 2 = (D 2 + E 2 - 4AF) / 4 A 2
Let's check to see that this works out for the example problem we set, i.e.
x 2 + y2 + 2x - 4y = 11
x 2 + y2 + 2x - 4y - 11 = 0
So: A = 1, D = 2, E = -4 and F = -11
r 2 = [2 2 + (-4) 2 - 4 (1)(-11))] / 4 2
r 2 = [4 + 16 + 44]/ 4 = 64/ 4 = 16
And since: r 2 = 16 then r = 4 which checks out.
The coordinate of the center will also be found to be:
h = (-D/ 2A) and k = (-E/ 2A)
Again, checking this for the example:
h = -2/ 2(1) = -1 and k = -(-4)/ 2 = 4/2 = 2
so (h, k ) = (-1, 2)
which again checks.
Problems for the math-minded (or readers looking for something to challenge them!)
1) Complete the square for the general analytic equation:
Ax 2 + Ay2 + Dx + Ey + F = 0
To show how the radius r is derived, as well as the coordinates for the center of a circle not necessarily at (0,0)
2) Check the equation: x 2 + y2 = 4
to see that the general form also applies here.
3) Find the coordinates of the center of each of the following circles and the radius r, also sketch the circle:
a) x 2 + y2 - 2 y = 3
b) 2x 2 + 2 y2 + x + y = 0
c) x 2 + y2 + 2x = 8
4) Obtain the equation for the circle shown below: