If $$a,b$$ are real numbers such that $$ab-4a+4b=18$$, find the minimum value of
$$a^2+b^2$$

Question

If $$a,b$$ are real numbers such that $$ab-4a+4b=18$$, find the minimum value of 
$$a^2+b^2$$

Uniteasy Admin · Accepted Answer

Let 
$$y = a^2+b^2$$n
$$x = a-b$$n
$$x^2-y = (a-b)^2 - (a^2+b^2) = -2ab$$
Then
$$ab = \dfrac{1}{2}(y-x^2) $$n
Onthe other hand,
$$ab-4a+4b=ab - 4x = 18$$
$$ab = 18+4x$$n
A quadratic equation is obtained from (3) and (4) 
$$ \dfrac{1}{2}(y-x^2) = 18+4x$$
$$y = x^2+8x+36 = (x+4)^2+20\geq 20 $$
which shows $$y_{min}$$ = 20 when $$x = -4$$
From (2) and (4), the following system of equations is obtained
$$ \begin{cases} a-b &= -4 \ ab &= 2 \end{cases}$$
Solving the system of equations gives
$$a = -2+\sqrt{6}, b = 2+\sqrt{6} $$  or $$a = -2-\sqrt{6}, b = 2-\sqrt{6} $$ 
Therefore,
The minimum value of $$a^2+b^2$$ is $$20$$

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