Question 1063040
 r = 2 ; (h, k) = (4, -4)
<pre>
{{{(x-h^"")^2+(y-k^"")^2}}}{{{""=""}}}{{{r^2}}}

Substitute 4 for h, -4 for k, and 2 for r

{{{(x-4^"")^2+(y-(-4)^"")^2}}}{{{""=""}}}{{{2^2}}}

Simplify,multiply everything out, collect terms, and get
0 on the right:

{{{(x-4^"")^2+(y+4^"")^2}}}{{{""=""}}}{{{4}}}

{{{(x-4^"")(x-4^"")+(y+4^"")(y+4^"")}}}{{{""=""}}}{{{4}}}

{{{(x^2-4x-4x+16^"")+(y^2+4y+4y+16)}}}{{{""=""}}}{{{4}}}

{{{x^2-4x-4x+16+y^2+4y+4y+16}}}{{{""=""}}}{{{4}}}

{{{x^2-8x+y^2+8y+32}}}{{{""=""}}}{{{4}}}

Get 0 on the right by subtracting 4 from both sides:

{{{x^2-8x+y^2+8y+28}}}{{{""=""}}}{{{0}}}

The general form has the squared terms first, then the
x and y terms then the constant (number), then = 0

{{{x^2+y^2-8x+8y+28}}}{{{""=""}}}{{{0}}}

Edwin</pre>