what do you mean with 'standard form?'

The form you want is probably (x-a)²+(y-b)²=r² because this is a circumference.

let me just show you a first approach to this type of transformations before moving on to your problem.

if you have an expression like this:
x²+px+q²
you can ALWAYS transform that into something like (x+z)²

x²+px+q²=(x+z)² ; p,q and zeR
x²+px+q²=x²+2zx+z²
px+q²=2zx+z²

now we only need to solve the system to find p and q
|px=2zx
|q²=z²


Now lets apply the above method.
x²+y²+9x-8y=0
(x²+9x+q1)+(y²-8y+q2)-q1-q2=0
lets work each square separately:

x²+9x+q1=(x+z)²
x²+9x+q1=x²+2xz+z²
9x+q1=2zx+z²

|9x=2zx  |z=9/2
|q1=z²    |q1=81/4

ok! we just found that
(x²+9x+q1)+(y²-8y+q2)-q1-q2=0
is equivalent to:
(x+9/2)²+(y²-8y+q2)-81/4-q2=0

Repeat the procedure for y.
y²-8y+q2=(y+z)²
y²-8y+q2=y²+2zy+z²
-8y+q2=2zy+z²

|-8y=2zy  |z=-4
|q2=z²      |q2=16

lets replace again in the first expression...
(x+9/2)²+(y-4)²-81/4-16=0
(x+9/2)²+(y-4)²-145/4=0
(x+9/2)²+(y-4)²=145/4

Remember also that the radius of this circumference is r²=145/4 => r=(√145)/2

-----------------------------
And here you have your 'standard form':
(x+9/2)²+(y-4)²=145/4

Last edited by kylekatarn (2005-07-29 12:27:31)

u r welcome : )