Expert: Josh - 10/27/2003

Question
What is the center and radius of the circle x^2+y^2-6x+8y=0

Answer
Iris,

Here is the question that I will pose to you.
What is the definition of a circle.

A circle is a locus, ie., a geometric curve on which every point is the same distance from a reference point, this, we refer to as the center.

How do we express distance in algebra or geometry?
Let x0 be our reference point.

In one dimension (1D), consider inequality expression such as |x-1|>=3.

This tells us that x is the set of points whose absolute distance is greater than or equal to three units from x=1.
The solution is found by consider the two cases,
CASE 1: If the term (x-1)>=0
(x-1)>3 gives x>4.
CASE 2: If the term (x-1)<0, then the absolute value |x-1| is equivalent to (1-x)
Solving |x-1|>=3 is the same as solving (1-x)>=3, which gives 1-3>=x. x<=-2

Draw this on a number line to see this visually.
......]____x____[.....

where ]=-2, x=1, [=4

---
Now, what happens in 2D?
This really boils down to how we define distance.
We use what we learnt from Pythagoras theorem, which states that given a right-angle triangle (ABC) as shown, where angle ^ABC=90 degrees, the length of the longest side,  (hypotenuse) squared, is equal to the sum of the squared distance of the other two sides.
A
|
|
B------------C

|AC|^2=|AB|^2+|BC|^2 ...[#1]
what do we see here?
|AC|,|AB|,|BC| are all absolute distances

Imagine point B being the center of the circle, in this special case, if point A and point C are to remain on the circle, then length|AB| must equal length |BC|. In fact, |AB|=|BC|=r, the radius of the circle.

Substituting this into [#1], we confirm that
|AC|^2=r^2+r^2=2*(r^2)
which is always true for a right angle triangle with two 45 degree angles. (After all, you may or may not have encountered this, but the first few trigonometric identities that you remember consist of sin(45)=cos(45)=1/sqrt(2), not to worry if you haven't seen these yet)

But the point is, the equation of a circle is not something mysterious. It is actually easy to remember.
r^2=(x-x0)^2+(y-y0)^2
where (x0,y0)= center of circle, r= radius.

So, the idea of the question is to get you to factorize the equation x^2+y^2-6x+8y=0 into this standard form.

Can you manage this?

x^2+y^2-6x+8y
=x^2-6x+y^2+8y ...collect like variables
=x^2-6x+y^2+8y ...complete the square (read notes below)
=x^2-3x-3x+9-9 +y^2+4y+4y+16-16
=(x^2-6x+9)-9 +(y^2+8y+16)-16
=(x^2-6x+9)+(y^2+8y+16)-25
=(x-3)^2+(y+4)^2-25
=0

Taking constant to RHS, yields
(x-3)^2+(y+4)^2=25
Refer to prior discussion,
center=(3,-4), radius=5
Watch out for the minus signs, don't get them confused.

|NOTES: in general x^2+ax
|=x^2+ax+(a/2)^2-(a/2)^2    ...add&subtract same quantity
|changes nothing
|=x^2+(a/2)x+(a/2)x+(a/2)^2-(a/2)^2 ...split middle term |into two halves
|=[x^2+2(a/2)x+(a/2)^2] -(a/2)^2 ...group quadratic,
recognize this
|=[x+(a/2)]^2 -(a/2)^2

I have said a lot of things, please revise this until you get the hang of it.

Cheers,
Josh