<p>In the standard (x,y) coordinate plane, what are the coordinates of the center of the circle whose equation is x^2-6x+y^2+4y+12=0
F. (-3, 2)
G. (-2,-3)
H. (-2, 3)
J. (3, -2)
K. (3, 2)</p>
<p>I know the answer because I have the key, but HOW do you get it??</p>
<p>x^2-6x+y^2+4y+12=0</p>
<p>x^2 - 6x + y^2 + 4y = -12</p>
<p>Complete the square for x’s and y’s</p>
<p>x^2 -6x + 9 + y^2 + 4y + 4 = -12 + 9 + 4
(x - 3)^2 + (y + 2)^2 = 1</p>
<p>(x - h)^2 + (y - k)^2 = r^2 is the equation of a circle with center (h,k). Thus:</p>
<p>(3, -2) D</p>
<p>Ah, duh! Thanks so much. I just couldn’t put the steps together in my head. Thanks</p>
<p>no problem</p>
<p>Okay I’ve got another.</p>
<p>Consider the equation sqrt b - sqrt a = 3 sqrt a, where a and b are positive real numbers. What is b in terms of a?</p>
<p>F. 16a
G. 9a
H. 4a
J. 3a
K. 2a</p>
<p>Could someone walk me through this, too?</p>
<p>sqrt b - sqrt a = 3 sqrt a</p>
<p>group like terms
so,</p>
<p>sqrt b = 4 sqrt a</p>
<p>square both sides</p>
<p>b = 16a</p>
<p>Wow I really need to think things through better. I understand all these tough concepts and then these totally easy problems stump me! Thanks for your help, I love this site</p>