<s>[INST] Proving a differential equation is a circle So, I have solved the differential equation, to find the general solution of: 
$$\frac{y^2}{2} = 2x - \frac{x^2}{2} + c$$
I am told that is passes through the point $(4,2)$. Using this information, I found $C$ to be $2$. Then I am asked to prove that the equation forms a circle, and find the radius and centre point. So I know to try and get it into the form: 
$$x^2 + y^2 = r^2$$
So I can get:
$$\frac{y^2}{2} + \frac{x^2}{2} = 2x + 2$$
Multiply both sides by $2$: 
$$y^2 + x^2 = 4x + 4$$
But from here I am not sure how to group the $x$'s to leave only a constant on the RHS. Any guidance would be much appreciated.
 [/INST] You have,
$$ y^2 +x^2 = 4x+4,$$
what we need to do is group the $x$'s together and complete the square, 
$$ y^2 + \left( x^2-4x \right) = 4.$$
So we need to complete the square for the expression $x^2-4x$. The idea is to write this polynomial as a perfect square plus a constant. To do this we first consider the expression $(x+a)^2 = x^2 + 2ax + a^2$. This motivates us to rewrite our polynomial as,
$$ x^2-4x = x^2-2(2)x  = (x-2)^2-2^2 = (x-2)^2-4$$
we can then substitute this back into the original equation to get,
$$ y^2 + \left( x-2\right)^2 = 8.$$
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