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kylekatarn
2005-07-29 12:44:35

u r welcome : )

Lily
2005-07-29 12:32:17

Thank you !

kylekatarn
2005-07-29 12:19:13

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

Lily
2005-07-29 11:53:39

the equation i just wrote is general forum (AX^2+Cy^2+Dx+Ey+F=0
i need convert standard form, the standard form for circle is
(x-h)^2+(y-k)^2=1
___        ___
a^2       b^2

kylekatarn
2005-07-29 11:44:43

what do you mean with 'standard form?'

Lily
2005-07-29 11:37:22

x^2+y^2+9x-8y=0

__________________
x^2+9x+(81/160)+y^2-8y+16=0
(x+9/4)^2+9y-40^2=(81/16)(16)
= (337/16)....
then i didn't know how to solve the rest of it..
plz help me ... thanx