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

Given equation of curves are
$xy = 4$ ….(i)

and ${x^2} + {y^2} = 8$ ……..(ii)

$\Rightarrow$ $x \cdot \frac{{dy}}{{dx}} + y = 0$

and $2x + 2y\frac{{dy}}{{dx}} = 0$

$\Rightarrow$ $\frac{{dy}}{{dx}} = \frac{{ - y}}{x}$

and $\frac{{dy}}{{dx}} = \frac{{ - 2x}}{{2y}}$

$\Rightarrow$ $\frac{{dy}}{{dx}} = \frac{{ - y}}{x} = {m_1}$

[say ]
and $\frac{{dy}}{{dx}} = \frac{{ - x}}{y} = {m_2}$ [say]

Since, both curves should have same slope.

Therefore,$\frac{{ - y}}{x} = \frac{{ - x}}{y} \Rightarrow - {y^2} = - {x^2}$

$\Rightarrow$ ${x^2} = {y^2}$ ….(iii)

Using the value of ${x^2}$ in Eq. (ii), we get
${y^2} + {y^2} = 8$

$\Rightarrow$ ${y^2} = 4 \Rightarrow y = \pm 2$

For $y = 2,x = \frac{4}{2} = 2$

and for $y = - 2,x = \frac{4}{{ - 2}} = - 2$

Therefore the required points of intersection are (2,2) and (-2,-2) .

For $(2,2),$ ${m_1} = \frac{{ - y}}{x} = \frac{{ - 2}}{2} = - 1$

and ${m_2} = \frac{{ - x}}{y} = \frac{{ - 2}}{2} = - 1$

${m_1} = {m_2}$

For $( - 2, - 2),$ ${m_1} = \frac{{ - y}}{x} = \frac{{ - ( - 2)}}{{ - 2}} = - 1$

and ${m_2} = \frac{{ - x}}{y} = \frac{{ - ( - 2)}}{{ - 2}} = - 1$

We can see that for both the intersection points slope of both the curves are same. Hence, the curves touch each other.