Prove that the curves $xy=4$ and $x^{2}+y^{2}=8$ touch each other.

(A) Given equations of the curves are $xy=4 \dots (i)$ and $x^{2}+y^{2}=8 \dots (ii)$.
Differentiating $(i)$ with respect to $x$: $x \frac{dy}{dx} + y = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$. Let this slope be $m_{1} = -\frac{y}{x}$.
Differentiating $(ii)$ with respect to $x$: $2x + 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y}$. Let this slope be $m_{2} = -\frac{x}{y}$.
For the curves to touch,they must intersect and have the same slope at the point of intersection. Setting $m_{1} = m_{2}$ gives $-\frac{y}{x} = -\frac{x}{y} \Rightarrow y^{2} = x^{2}$.
Substituting $x^{2} = y^{2}$ into $(ii)$: $y^{2} + y^{2} = 8 \Rightarrow 2y^{2} = 8 \Rightarrow y^{2} = 4 \Rightarrow y = \pm 2$.
If $y = 2$,then $x = \frac{4}{2} = 2$. If $y = -2$,then $x = \frac{4}{-2} = -2$. The points of intersection are $(2, 2)$ and $(-2, -2)$.
At $(2, 2)$: $m_{1} = -\frac{2}{2} = -1$ and $m_{2} = -\frac{2}{2} = -1$. Since $m_{1} = m_{2}$,the curves touch at $(2, 2)$.
At $(-2, -2)$: $m_{1} = -\frac{-2}{-2} = -1$ and $m_{2} = -\frac{-2}{-2} = -1$. Since $m_{1} = m_{2}$,the curves touch at $(-2, -2)$.
Thus,the curves touch each other.