write the given equation in standard form: x^2+y^2-4x-6y-13=0

Question

anonymous · Answer

How about (x-2)^2 + (y-3)^2 = 26 http://www.oojih.com for free study guide and review questions

anonymous · Answer

do you know how to 'complete the square"?

anonymous · Answer

first rewrite as 
$$x^2-4x+y^2-6x=13$$ then half of -4 is -2 and half of -6 is -3 so write as 
\[(x-2)^2+(y-3)^2=13+2^2+3^2=13+4+9=26

anonymous · Answer

$$(x-2)^2+(y-3)^2=13+2^2+3^2=13+4+9=26$$

watchmath · Answer

$(x-2)^2+(y-3)^2=13+4+9$
$(x-2)^2+(y-2)^2=30$
$\frac{(x-2)^2}{30}+\frac{(y-3)^2}{30}=1$

watchmath · Answer

lol you were correct satellite

anonymous · Answer

hi watchman

anonymous · Answer

yeah arithmetic is tough this time of night!

watchmath · Answer

still energetic answering questions :)?

anonymous · Answer

so is typing "watchmath" instead of "watchman"  sorry

anonymous · Answer

thanks guys, very helpful! I appreciate it!

anonymous · Answer

bb911 is it clear how to complete a square?

anonymous · Answer

yes, I understand now, thanks!