Finding the minimum

Let $x = r \cos {\theta}$ and let $y = r \sin {\theta}.$

Now we need to find the minimum value of $x^2 + y^2,$ which simplifies to $r^2.$

The given equation, on substitution, simplifies to $\frac{r^2}{4} \cdot \sin ( 4\theta ) = 1,$ which implies $r^2 = 4 * \text{csc} ( 4\theta).$

Hence, the minimum value of $r^2 = 4.$

Applying A.M.-G.M. inequality-

$(x^2-y^2)^2 + 4 x^2y^2 \geq 2 \sqrt{4 x^2y^2(x^2-y^2)^2}=4 xy(x^2-y^2)=4(x^2+y^2)) \\ \implies (x^2+y^2)^2 \geq 4(x^2+y^2) \\ \implies (x^2+y^2) \geq 4$

Minimum value is $4$ occurs when $x^2-y^2=|2xy|$

Using the AM-GM inequality,

$\large \frac{(x^2 - y^2) + (x^2 - y^2) + (2 \sqrt{2} + 2)y^2 + (2 \sqrt{2} - 2)x^2 }{4} \geq \sqrt[4]{(x^2 - y^2)^2x^2y^2(2 \sqrt{2} + 2)( 2 \sqrt {2} - 2)}$ .

Simplifying,

$2 \sqrt{2} (x^2 + y^2) \geq 4 \sqrt{2xy(x^2 - y^2)} \implies (x^2 + y^2) \geq 2 \sqrt{xy(x^2 - y^2)}$

Squaring both sides and dividing by $x^2 + y^2$ , we have

$x^2 + y^2 \geq 4(\frac{xy(x^2 - y^2)}{x^2 + y^2}) = 4$ ,

Giving us

$x^2 + y^2 \geq \boxed{4}$