**Hint** $\ $ The transformation is simply *multiplication by* $\, 1+i\,$ in the complex plane

$$(1+i)(x+yi)\, =\, x\!-\!y + (x\!+\!y)i $$

Thus if $\ |x+yi| = x^2+y^2 = c \ $ then

$$ (x\!-\!y)^2+(x\!+\!y)^2 =\, |x\!-\!y+(x\!+\!y)i| = |(1+i)(x+yi)|\, =\, 2\,|x+yi|\,=\,2c$$

Geometrically, multiplication by $\ 1+i = \sqrt{2}\, e^{\large i\pi/4}\,$ may be visualized as an expansion by the factor $\,\sqrt{2}\,$ composed with a rotation by $\,\pi/4.\,$ Let's visualize this in the diagram from Oleg567's answer, excerpted below. Let $\,\rm\color{#c00}{v = x+y\, i}\,$ be the lower red vector. Multiplying it by $\,1+i\,$ yields $\, (1+i){\rm v = v} + i\rm v,\,$ where $\,\rm\color{#c00}{i\,v = -y + x\, i}\,$ is the rotation of $\,\rm v\,$ by $\,\pi/4.\,$ Adding these red vectors yields the result $\,\rm\color{blue}{v + i\, v = x-y + (x+y)\,i}.$

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