A conic is a shape obtained from taking a plane slice through a double cone. Below are all the different types of shapes possible.
[DIAGRAM OF CONIC SHAPES]
Remark: A circle is just a special case of an ellipse. Also notice that ellipses and hyperbolas have centers; that is, there exists a point $C$ such that a rotation about $C$ through an angle $\pi$ is a symmetry of the conic.
We define the set of conic sections that are ellipses, parabolas, or hyperbolas to be non-degenerate conic sections and we define the set of conic sections that are single points, single lines, or pairs of lines to be degenerate conic sections.
Circles
Think back to when we were all learning how to draw circles. One technique our teachers might’ve taught us is to attach one end of a string to a piece of paper and a pencil to the other. Then, with the string stretched outwards, a perfect circle can be drawn with the pencil.
This demonstrates a fundamental property about circles: a circle in $\R^2$ is the set of all points $(x,y)$ that lie a fixed distance, called the radius, away from a fixed point, called the center.
Recall that the distance between two points $p_1=(x_1, y_1)$ and $p_2=(x_2, y_2)$ in $\R^2$ is given by the formula $\text{dist}(p_1,p_2) = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}.$
Then for any circle with center $C=(a, b)$, we have that any arbitrary point, $(x,y)$, on the circle is given by \[r^2=(x-a)^2+(y-b)^2\] where $r$ is the radius.
Example 1
Determine the equation of each of the circles with the following center and radius:
(a) center: origin, radius: $3$
(b) center: $(4,2)$, radius: $7$
(c) center: $(\sqrt{2}, 3)$, radius: $\sqrt[3]{7}$
Solution
Here, we're just plugging numbers into the formula we found earlier:
(a) $(x-0)^2 + (y-0)^2 = 3^2 \implies x^2+y^2=9$
(b) $(x-4)^2 + (y-2)^2 = 49$
(c) Note that $\sqrt[3]{7}=7^{\frac{1}{3}}$. So then $(x-\sqrt{2})^2 + (y-3)^2 = 7^{\frac{2}{3}}$
If we expand our formula, we get that \[x^2 + y^2 -2ax -2by + (a^2 +b^2 -r^2) = 0.\] So then let $f=-2a$, $g=-2b$, and $h=a^2 +b^2 -r^2$. Our equation becomes \[x^2+y^2+fx+gy+h=0.\]
Example 2 Determine the condition on the numbers $f,g,$ and $h$ in the equation \[x^2+y^2+fx+gy+h=0.\] for the circle with this equation to pass through the origin.
Solution Since all points on the circle satisfy the given equation, $(0,0)$ must be a solution if our circle contains the origin. Plugging in $0$ for $x$ and $y$, we find that $h=0$. But recall that $h=a^2+b^2-r^2$. Thus, we obtain the three following conditions that $f,g,$ and $h$ must satisfy:
\[f=-2a\] \[g=-2b\] \[h=0 \implies r^2=a^2+b^2\]
So we’ve seen that the equation for a circle can be written as $x^2 + y^2 +fx +gy + h = 0.$ But can we go the other way? In other words, given an equation of the second form, can we determine if it represents a circle and find its center and radius?
Recall completing the square. Isolate terms that involve only $x$ and terms that involve only $y$. Complete the squares of these two expressions and plug back into the original equation. For example, consider \[x^2+y^2-2x-4y+2=0.\] Writing only the terms that involve $x$ and the terms that involve $y$ and completing the square, we obtain \[x^2-2x = (x-1)^2-1\] \[y^2-4y=(y-2)^2-4.\]
Plugging this back into our original equation, we get
\begin{align*} (x-1)^2-1+(y-2)^2-4+2 &= 0 \\ (x-1)^2+(y-2)^2 &= 3 \end{align*}
We can now easily see that the center is $(1,2)$ and the radius is $\sqrt{3}$.
Similarly, if we apply the same method to the general form $x^2 + y^2 +fx +gy + h = 0,$ we obtain the equation \[\left(x+\frac{1}{2}f\right)^2 +\left(y+\frac{1}{2}g\right)^2 = \frac{1}{4}f^2+\frac{1}{4}g^2-h.\]
Theorem 1 An equation of the form \[x^2 + y^2 +fx +gy + h = 0\] represents a circle with
center $(-\frac{1}{2}f, -\frac{1}{2}g)$ and radius $\sqrt{\frac{1}{4}f^2+\frac{1}{4}g^2-h},$
provided that $\frac{1}{4}f^2+\frac{1}{4}g^2-h > 0.$
Example 3 Determine the center and the radius of the following equation: \[4x^2+4y^2-12x-40y=0.\]
Solution Since the coefficient in front of the $x^2$ and $y^2$ term is not $1$, divide the equation by $4$ to obtain \[x^2+y^2-3x-10y=0.\] Note that $f=-3,$ $g=-10,$ and $h=0.$ By theorem $1,$ we have that this equation describes a circle with center at $(\frac{3}{2}, 5)$ and radius $\frac{\sqrt{109}}{2}.$
Orthogonal Circles
Here, we pose the question: under what conditions will two intersecting circles intersect at right angles?
Consider a circle $C_1$ with center $A = (-\frac{1}{2}f_1, -\frac{1}{2}g_1)$ and radius $r_1=\sqrt{\frac{1}{4}f_1^2 + \frac{1}{4}g_1^2-h_1}$ and a circle $C_2$ with center $B = (-\frac{1}{2}f_2, -\frac{1}{2}g_2)$ and radius $r_2=\sqrt{\frac{1}{4}f_2^2 + \frac{1}{4}g_2^2-h_2}.$ Let $P$ be one of their points of intersection and examine the triangle $\triangle ABP.$
[DIAGRAM OF INTERESECTING CIRCLES]
If the circles intersect orthogonally, then the line $AP$ is tangent to $C_2$ which makes it orthogonal to $PB.$ Thus, $\triangle ABP$ is a right triangle and we can apply the Pythagorean Theorem to obtain
\begin{equation}
\label{eq:orthog}
(AP)^2 + (PB)^2 = (AB^2)
\end{equation}
But $AP$ and $PB$ are just the radii of their respective circles. Thus, we know their values:
\begin{align*}
(AP)^2 &= r_1^2 = \frac{1}{4}f_1^2 + \frac{1}{4}g_1^2-h_1 \;\text{ and} \\
(PB)^2 &= r_2^2 = \frac{1}{4}f_2^2 + \frac{1}{4}g_2^2-h_2.
\end{align*}
And using the distance formula,
\begin{align*}
(AB)^2 &= (\frac{1}{2}f_1 - \frac{1}{2}f_2)^2 + (\frac{1}{2}g_1 - \frac{1}{2}g_2)^2 \\
&=(\frac{1}{4}f_1^2- \frac{1}{2}f_1f_2 + \frac{1}{4}f_2^2) + (\frac{1}{4}g_1^2- \frac{1}{2}g_1g_2 + \frac{1}{4}g_2^2)
\end{align*}
Substituting these values into equation \ref{eq:orthog} and simplifying, we find that
\begin{align*}
-h_1 - h_2 &= -\frac{1}{2}f_1f_2 - \frac{1}{2}g_1g_2 \\
\implies f_1f_2 + g_1g_2 &= 2(h_1 + h_2)
\end{align*}
Thus, we arrive at our second theorem:
Theorem 2 Two intersecting circles $C_1$ and $C_2$ with equations \begin{align*} x^2 + y^2 +f_1x +g_1y + h_1 &= 0 \;\text{ and} \\ x^2 + y^2 +f_2x +g_2y + h_2 &= 0 \end{align*} respectively, are orthogonal if and only if \[f_1f_2 + g_1g_2 = 2(h_1 + h_2).\]
Proof See the above discussion.
Circles Through Two Points
Suppose now we wish to find the set of all circles that pass through two given points. Let $C_1$ and $C_2$ be circles with equations as before and suppose that they intersect at points $P$ and $Q.$ Then, if $k\neq -1,$ \begin{equation} \label{eq:circtwopoints} x^2 + y^2 +f_1x +g_1y + h_1 +k(x^2 + y^2 +f_2x +g_2y + h_2) = 0 \end{equation} represents a circle that passes through $P$ and $Q.$ To convince yourself of this fact, note that the above equation can be turned into a general form for a circle for any value of $k.$ Additionally, the left side equals $0$ because of the general definition of a circle for $C_1$ and $C_2.$
If $k=-1,$ then equation \ref{eq:circtwopoints} represents a line that passes through $P$ and $Q.$
Remark We can consider the circle $C_2$ to be the $k=\infty$ case. To see this, divide equation \ref{eq:circtwopoints} by $k$ and let $k\rightarrow\infty.$
Theorem 3 Let $C_1$ and $C_2$ be circles with equations \begin{align*} x^2 + y^2 +f_1x +g_1y + h_1 &= 0 \; \text{ and} \\ x^2 + y^2 +f_2x +g_2y + h_2 &= 0 \end{align*} that intersect at distinct points $P$ and $Q.$ Then the line and any circle (other than $C_2$) through $P$ and $Q$ have an equation of the form \[x^2 + y^2 +f_1x +g_1y + h_1 +k(x^2 + y^2 +f_2x +g_2y + h_2) = 0\] for some number $k\in\R.$
If $k\neq -1,$ this is an equation for a circle; if $k=-1,$ this is the equation of a line.
Focus-Directrix Definition of the Non-Degenerate Conics
So far in this section, we’ve explored circles by defining it as a set of points at a fixed distance from a fixed point. Now, we wish to explore the other non-degenerate conics. These can be defined as a set of points $P$ that is a distance away from a fixed point, which we call its focus. This distance from the focus is a constant multiple, which we call its eccentricity $(e)$, of the distance from a fixed line, which we call its directrix.
Eccentricity A non-degenerate conic is an ellipse if $0\leq e < 1,$ a parabola if $e=1,$ or a hyperbola if $e>1.$