Let $$a,b$$ be real numbers, such that $$ab-4a+4b=18$$. Find the minimum value of $$a^2+b^2$$

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Uniteasy Admin · Accepted Answer

Let 
$$x = a-b$$nc
$$y = a^2+b^2$$nc
Taking square of equation (1) gives 
$$x^2 =a^2+b^2-2ab$$nc
Substitute (2) and solve for $$ab$$
$$ab = \dfrac{y-x^2}{2} $$nc
Substitute (4) and(1) to given equation $$ab-4a+4b=18$$, then
$$\dfrac{y-x^2}{2}-4x = 18$$c
Rearrange terms to the form of a quadratic function.
$$\begin{aligned} y&= x^2+8x+36 \ &= (x+4)^2+20 \end{aligned}$$c
Therefore, the minimum value of $$y$$, that is, $$ a^2+b^2$$ is $$20$$ when $$x= -4$$

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