General form of a circle

Question

My math teacher taught me that the general form (equation) of a circle is:

$$ ax^2+by^2+cx+dy+e=0 $$

He also asked us this:

If the product of $c$ and $d$ is negative, then what 2 quadrants can the graph be in? I am really confused on how each of the coefficients change the graph. For example, if $e$ is negative, the graph would move to the left and reflect across the $y$ axis, so it would be in quadrant $2$ (This is just an example I know I am wrong). Would you please help me out by telling me what each of the coefficients ($a$, $b$, $c$, $d$, and $e$) do to the graph if it is negative or positive (including which quadrant the graph of the circle would be in)? If you could be as detailed as possible, I would love it. Thank you so much for the help, I appreciate it!

Answer 1

For a circle, we may assume that $a=b$. Now $$ \begin{align*} & bx^2+by^2+cx+dy+e=0 \\ \Rightarrow & x^2+\frac{c}{b}x+y^2+\frac{d}{b}y=-\frac{e}{b}\\ \Rightarrow & x^2+\frac{c}{b}x+\left(\frac{c}{2b}\right)^2+y^2+\frac{d}{b}y+\left(\frac{d}{2b}\right)^2=\left(\frac{c}{2b}\right)^2+\left(\frac{d}{2b}\right)^2-\frac{e}{b}\\ \Rightarrow & \left( x+\frac{c}{2b} \right)^2+\left( y+\frac{d}{2b} \right)^2=\frac{c^2+d^2-4be}{4b^2}. \end{align*} $$

Certain things come to light in this form. Firstly, the radius is $$r=\sqrt{\frac{c^2+d^2-4be}{4b^2}},$$ but we require $c^2+d^2-4be>0$, and $b \neq 0$ if we are to have real non-zero radius.

The center is clearly at $\displaystyle (x,y)=\left( -\frac{c}{2b},-\frac{d}{2b} \right)$.

You might fix some values then vary the others to see what they do. I made this dynamic slider graph for you. Start by fixing $b$ and $e$, then vary $c$ and $d$ to see what happens. Try some other experiments after you have that effect. For example, fix everything else, then vary $e$. After substantial experimentation with the graph, and time pondering your equation, I am quite sure you will be enlightened.

Answer 2

For a circle, you must have $a=b$. Otherwise if it is closed the curve will be an ellipse. Divide by $a$ (if it is zero you have a straight line) to get $x^2+y^2+c'x+d'y+e'=0$ (where the primes indicate they are different values from before). Complete the squares: $(x+\frac {c'}2)^2+(y+\frac{d'}2)^2=\frac 12(c'^2+d'^2)-e'$ The center is at $(-\frac {c'}2,-\frac {d'}2)$ and the radius is $\sqrt{\frac 12(c'^2+d'^2)-e'}$. If the stuff under the square root sign is negative, you have an empty set.

Answer 3

Usually we define circle to be:

Locus of a point having fixed distance $r$ from a point $(p,q)$

We usually say that distance to be "Radius" and that point $(p,q)$ to be center.

distance from an arbitrary point to $(p,q)$ is $\sqrt{(x-p)^2+(y-q)^2}$

We need this to be equal to $r$

i.e., $\sqrt{(x-p)^2+(y-q)^2}=r \Rightarrow (x-p)^2+(y-q)^2=r^2$

i.e., $x^2+y^2-2px-2qy+p^2+q^2=r^2$ may be we like to see it as :

$x^2+y^2-2px-2qy+p^2+q^2-r^2=0$ May be we like to see it as :

$x^2+y^2+lx+my+n=0$ where $l=-2p~; ~m=-2q ~;~ n=p^2+q^2-r^2$

i.e., center $(x,y)$ is $(\frac{-l}{2},\frac{-m}{2})$

Once we know what is $c$ then radius is obtained as : $r=\sqrt{n^2-p^2-q^2}$

Your general form $ax^2+by^2+cx+dy+e=0$ does not represent a circle is should be

$ax^2+ay^2+cx+dy+e=0$ which can be seen as $x^2+y^2+\frac{c}{a}x+\frac{d}{a}y+\frac{e}{a}=0$

We equate this to $x^2+y^2+lx+my+n=0$ to conclude that :

Center is given by $(\frac{-l}{2},\frac{-m}{2})=(\frac{-c}{2a},\frac{-c}{2a})$ and radius is $r=\sqrt{n^2-p^2-q^2}=\sqrt{(\frac{e}{a})^2-(\frac{-c}{2a})^2-(\frac{-c}{2a})^2}$

So, For general equation of circle $ax^2+ay^2+cx+dy+e=0$ have :

• Center $(\frac{-c}{2a},\frac{-c}{2a})$
• Radius $\sqrt{(\frac{e}{a})^2-(\frac{-c}{2a})^2-(\frac{-c}{2a})^2}$

So, depending upon sign of $(\frac{-c}{2a},\frac{-c}{2a})$ and value of $\sqrt{(\frac{e}{a})^2-(\frac{-c}{2a})^2-(\frac{-c}{2a})^2}$ we may know what is the location of the circle.

I guess you are familiar with :

• $(+,+)$ first Quadrant
• $(-,+)$ second Quadrant
• $(-,-)$ third Quadrant
• $(+,-)$ fourth Quadrant

So, If $(\frac{-c}{2a},\frac{-c}{2a})$ is of the form $(+,+)$ then center is located in first Quadrant and similar things can be said about the center depending upon other cases.

Radius is what assures the containment which can be said by first look in contrast with position of center.

It is not necessary that circle is totally contained in only a particular quadrant.