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**Lily****Guest**

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

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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

**kylekatarn****Member**- Registered: 2005-07-24
- Posts: 445

what do you mean with 'standard form?'

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**Lily****Guest**

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****Member**- Registered: 2005-07-24
- Posts: 445

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 (2005-07-28 14:27:31)*

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**Lily****Guest**

Thank you !

**kylekatarn****Member**- Registered: 2005-07-24
- Posts: 445

u r welcome : )

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