Show that the curves $xy=a^2$ and $x^2+y^2=2a^2$ touch each other.
The two curves are
$xy=a^2$ $\implies$ $x = \dfrac{a^2}{y}$ $\;\; \cdots$ (1)
$x^2 + y^2 = 2a^2$ $\implies$ $\left(\dfrac{a^2}{y}\right)^2 + y^2 = 2a^2$ [in view of equation (1)]
i.e. $a^4 + y^4 = 2a^2 y^2$
i.e. $y^4 - 2a^2 y^2 +a^4 =0$
i.e. $\left(y^2-a^2\right)^2 = 0$
i.e. $y^2 -a^2 =0$ $\implies$ $y^2 = a^2$ $\implies$ $y = \pm a$
$\therefore$ $\;$ From equation (1),
when $y = a$, $x=a$ and
when $y = -a$, $x=-a$
$\therefore$ $\;$ The given curves intersect each other at the points $\left(a,a\right)$ and $\left(-a,-a\right)$
Differentiating $xy=a^2$ w.r.t x gives
$x \dfrac{dy}{dx}+y=0$
i.e. $\dfrac{dy}{dx}=\dfrac{-y}{x}$
$\therefore$ $\;$ $\dfrac{dy}{dx}\bigg|_{\left(a,a\right)}= \dfrac{-a}{a}=-1=m_1 \left(\text{say}\right)$ $\;\; \cdots$ (2)
Differentiating $x^2 + y^2 = 2a^2$ w.r.t x gives
$2x+2y\dfrac{dy}{dx}=0$
i.e. $\dfrac{dy}{dx}=\dfrac{-x}{y}$
$\therefore$ $\;$ $\dfrac{dy}{dx}\bigg|_{\left(a,a\right)}= \dfrac{-a}{a}=-1 = m_2 \left(\text{say}\right)$ $\;\; \cdots$ (3)
Let $\phi$ be the angle between the two tangents.
Then $\tan \phi = \left|\dfrac{m_1 - m_2}{1+m_1 m_2}\right| = \left|\dfrac{-1+1}{1+\left(-1\right)\left(-1\right)}\right|=0$
$\implies$ $\phi = 0$
i.e. the two tangents touch each other.
$\implies$ The two curves touch each other.