Question 426851
<font face="Times New Roman" size="+2">


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ h)^2\ +\ (y\ -\ k)^2\ =\ r^2]


is the equation of a circle with center at *[tex \Large (h,k)] and radius *[tex \Large r]


Move the constant to the RHS


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 18x\ +\ y^2\ +\ 4y\ =\ -4]


Divide the coefficient on the first degree *[tex \Large x] term by 2 and square the result.  Add that result to both sides, completing the square in *[tex \Large x] .  Do the same thing with the coefficient on the first degree *[tex \Large y] term.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 18x\ +\ 81\ +\ y^2\ +\ 4y\ +\ 4\ =\ -4\ +\ 81\ +\ 4]


Collect terms in the RHS and factor the two perfect square trinomials in the LHS.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 9)^2\ +\ (y\ +\ 2)^2\ =\ 81]


Finally, rewrite your equation thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 9)^2\ +\ \left(y\ -\ (-2)\right)^2\ =\ 9^2]


Now you can determine the center and radius by inspection.  Compare to the pattern given at the beginning.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>