# Standard form!

• Nov 16th 2006, 11:47 AM
cutie4ever
Standard form!
okay so the question is as follows!
use X^2+ y^2-6x+8y+ 9= 0
write the equation in standard form!

so here is waht I have done!

X^2+ y^2 -6x +8y+ 9= 0

X^2+ y^2 -6x +8y=-9

now I have to remove the coefiicent from X^2 and y^2
1(X^2-6x) +1( y^2 +8y) =-9

Now I have to complete the squares but I get lost from here I am unsure how to do it! thanks! :)
• Nov 16th 2006, 12:22 PM
topsquark
Quote:

Originally Posted by cutie4ever
okay so the question is as follows!
use X^2+ y^2-6x+8y+ 9= 0
write the equation in standard form!

so here is waht I have done!

X^2+ y^2 -6x +8y+ 9= 0

X^2+ y^2 -6x +8y=-9

now I have to remove the coefiicent from X^2 and y^2
1(X^2-6x) +1( y^2 +8y) =-9

Now I have to complete the squares but I get lost from here I am unsure how to do it! thanks! :)

If you have \$\displaystyle x^2 + 2ax\$ then you add and subtract \$\displaystyle a^2\$:

\$\displaystyle x^2 - 6x\$:
\$\displaystyle 2a = -6\$, so \$\displaystyle a = -3\$ so \$\displaystyle a^2 = 9\$:

\$\displaystyle (x^2-6x + 9 - 9) +(y^2 +8y) =-9\$

\$\displaystyle (x^2 - 6x + 9) - 9 + (y^2 +8y) =-9\$

\$\displaystyle (x - 3)^2 + (y^2 +8y) = 0\$

Similarly:
\$\displaystyle (x - 3)^2 + (y^2 + 8y + 16 - 16) = 0\$

\$\displaystyle (x - 3)^2 + (y^2 + 8y + 16) - 16 = 0\$

\$\displaystyle (x - 3)^2 + (y + 4)^2 = 16\$

\$\displaystyle (x - 3)^2 + (y + 4)^2 = 4^2\$

So the equation is for a circle of radius 4 with center at (3, -4).

-Dan