# Standard form!

• November 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! :)
• November 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 $x^2 + 2ax$ then you add and subtract $a^2$:

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

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

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

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

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

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

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

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

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

-Dan