Question

Given the general form of the equation of a circle, x^2+y^2+ax+by+c=0, explain how to (i) write the equation in standard form, and (ii) how to graph the circle in standard form. Also demonstrate (i) and (ii) by providing an example. Please show work.

Answer 1

group x terms together
and
group y terms together
and
subtract constant term from both sides
\[x^2+ax+y^2+by=-c\]
now we will complete the square for the x part
and
we will complete the square for the y part
(don't forget whatever you add on one side you must add to the other)
\[x^2+ax+(\frac{a}{2})^2+y^2+by+(\frac{b}{2})^2=-c+(\frac{a}{2})^2+(\frac{b}{2})^2\]
\[(x+\frac{a}{2})^2+(y+\frac{b}{y})^2=-c+\frac{a^2}{4}+\frac{b^2}{4}\]
\[(x+\frac{a}{2})^2+(y+\frac{b}{2})^2=\frac{-4c}{4}+\frac{a^2}{4}+\frac{b^2}{4}\]
\[(x+\frac{a}{2})^2+(y+\frac{b}{2})^2=\frac{a^2+b^2-4c}{4}\]
\[\text{ center is } (\frac{-a}{2},\frac{-b}{2})\]
\[\text{ radius is } \sqrt{\frac{a^2+b^2-4c}{4}} \text{ if } a^2+b^2-4c \ge 0\]

Answer 2

Well, I'm kind of making an educated guess here.
The standard equation of a circe is
\[(x-h)^2+(y-k)^2=r^2\], for a circle centered at (h, k) and of radius r.

If you start from
\[x^2+y^2+ax+by+c=0\],
you'd have to complete the squares in order to reach the standard form.

You first reorganize the terms,
\[(x^2+ax)+(y^2+by)=-c\]
And now you complete squares. You have to have in mind that
\[(x+d)^2=x^2+2dx+d^2)\]
And with the general equation you have part of the expansion of the squared binomial (which is called perfect square trinomial). You have to add something to those partial expansions in such a way that the become a perfect square trinomial. And how to do it? It's easy.
If you have, say
\[x^2+ax\]
You'd have to add AND SUBTRACT (this is very important for the equation to still be valid) the following term:
\[(a/2)^2\]
Now, you had
\[(x^2+ax)+(y^2+by)=-c\]
You have to add and subtract (a/2)^2 and (b/2)^2
\[(x^2+ax+(a/2)^2)-(a/2)^2+(y^2+by+(b/2)^2)-(b/2)^2=-c\]
Now you have two perfect square trinomials, which can be factorized,
\[(x+(a/2))^2+(y+(b/2))^2=(a/2)^2+(b/2)^2-c\]
Note that I moved the subtractions to the right side of the equation.

Now you have a circumference in the standard form. This one in particular is centered at (-(a/2),-(b/2)) and has a radius of \[\sqrt{(a/2)^2+(b/2)^2-c}\].

Now, about graphing the circle. It's easy. In the standard form, your circle is centered at (h,k). You also have your radius r. So just take a compass, center it at (h,k), and trace the circumference with a radius r.

Answer 3

Thank you myininaya!

Answer 4

Is there any way you can draw the graph ? I have no idea how to do it.

Answer 5

Thank you mrreevers