Circles are present in our everyday lives - the bottoms of our mugs, buttons on our TV remotes and the pizza that we eat. Rarely do we consider how these objects are created; the work placed into making sure your cookie cutter is in perfect shape. The secret? MATH!
Before we begin graphing, let us review some key terms that are crucial to our understanding:
Circle: A two-dimensional geometric figure, consisting of the set of all those points in a plane that are equally distant from another point.
Radius: A line segment between any point on the circumference of a circle and its center.
Diameter: Two times the radius of a circle.
Area: The interior surface of a circle, given by \( A = \pi r^2 \)
Equation of a Circle
As briefly discussed above, a circle is defined as the set of points that are a fixed distance from a center point. The distance formula we use in the Pythagorean Theorem can be extended directly to this definition by noting that the radius is the distance between the center of a circle and its edge. We now have the general equation for a circle centred at \( (a, b) \) with radius \( r \) and points \( (x, y) \) to be:
\[ (x-a)^2 + (y-b)^2 = r^2 \]
If the circle is centred at the origin, \( (0, 0) \), then its equation is: \[ x^2 + y^2 = r^2 \]
Consider sketching the graph of \( x^2 + y^2 = 16 \).
We first find the center of the circle
\( \Longrightarrow \) in this case it is the origin, \( (0, 0) \)
Next, we determine the radius \( \Longrightarrow r^2 = 16 \)
\( \Longrightarrow r = \sqrt{16} \)
\( \Longrightarrow r = 4 \)
We reject the negative square root since we are dealing with distance. Our final step to this question is to plot the radius points [\( (0, 4), (0, -4), (4, 0), (-4, 0) \) on the coordinate plane and connect the dots using a round curve:
What if the circle is not centred at the origin?
EXAMPLE
Sketch the graph of \( (x-3)^2 + (y +1)^2 = 25 \).
See the video below for the solution:
Try this interactive tool!
Adjust the sliders for \(a\), \(b\), and \(r\). Observe what happens to the circle.
What happens when our circle is not perfectly symmetrical? Can we distort the shape to make it seem stretched our compressed? The answer is yes!
Ellipses are, in simplest terms, distorted circles. They share many of the same attributes, but the key differences are their equations and graphs.
The standard form for the equation of an ellipse with center \( (0, 0) \) and major axis on the \(x-axis \) is
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Where
1. \( a > b \)
2. the length of the major axis is \( 2a \)
3. the length of the minor axis is \( 2b \)
4. the coordinates of the "vertices" are \( (a, 0) \)
5. the coordinates of the co-vertices are \( (0, b) \)
6. the coordinates of the foci are \( (c, 0) \) ; \( c^2 = a^2 - b^2 \)
The standard form for the equation of an ellipse with center \( (0, 0) \) and major axis on the \(y-axis \) is
\[ \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \]
Where
1. \( a > b \)
2. the length of the major axis is \( 2a \)
3. the length of the minor axis is \( 2b \)
4. the coordinates of the "vertices" are \( (0, a) \)
5. the coordinates of the co-vertices are \( (b, 0) \)
6. the coordinates of the foci are \( (0, c) \) ; \( c^2 = a^2 - b^2 \)
The stretch/compression of an ellipse is its eccentricity and is given by the formula \( e = \frac{c}{a} \).
Try this interactive tool!
Adjust the sliders for \(a\) and \(b\) and observe what happens. The points \(P\) and \(Q\) represent the foci of the ellipse. Notice that the foci switch axis depending on whether the ellipse is stretched horizontally (\(a > b\)) or stretched vertically (\(b > a\)).
Now that we know our basic properties of ellipses, let us apply our knowledge to graph these shapes.
Determine the standard form of an ellipse that has vertices \( (8, 0) \) and foci \( (5, 0) \). Then draw the ellipse.
Solution:
\( \Longrightarrow \) the foci are on the \( x-axis \), and thus that is our major axis and \( a = \pm 8 \)
\( \Longrightarrow a^2 = (\pm 8)^2 = 64 \) and \( c^2 = (\pm 5)^2 = 25 \)
We can now use our equation \( c^2 = a^2 - b^2 \) to solve for b:
\( \Longrightarrow 25 = 64 - b^2 \)
\( \Longrightarrow b^2 = 64-25 \)
\( \Longrightarrow b^2 = 39 \longrightarrow b = \pm \sqrt{39} \)
Our ellipse is represented by the equation \( \frac{x^2}{64} + \frac{y^2}{39} = 1 \) and has the following graph:
[image here]
EXAMPLES
a) Find the equation of an ellipse whose vertices are \( (2, -2) \) and \( (2, 4) \) and whose eccentricity is \( \frac{1}{3} \).
b) Graph the ellipse.
See the video below for the solution: