Prove that the curves xy = 4

Given curves are equations of two circles,

xy = 4 ….. (i) and

x2 + y2 = 8 …. (ii)

Different equation (i) w.r.t., x

$x \cdot \frac{d y}{d x}+y \cdot 1=0$

$\Rightarrow \frac{d y}{d x}=-\frac{y}{x} \Rightarrow m_{1}=-\frac{y}{x}$ $\ldots($ iii $)$

where, $m_{1}$ is the slope of the tangent to the curve. Differentiating eq. (ii) w.r.t. $x$

$2 x+2 y \cdot \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\frac{x}{y} \Rightarrow m_{2}=-\frac{x}{y}$

where, $m_{2}$ is the slope of the tangent to the circle. To find the point of contact of the two circles

$m_{1}=m_{2} \Rightarrow-\frac{y}{x}=-\frac{x}{y} \Rightarrow x^{2}=y^{2}$

Putting the value of $y^{2}$ in eq. (ii)

$x^{2}+x^{2}=8 \Rightarrow 2 x^{2}=8 \Rightarrow x^{2}=4$

Thus, $\quad x=\pm 2$

And, $\quad x^{2}=y^{2} \Rightarrow y=\pm 2$

Therefore, the point of contact of the two circles are (2, 2) and (-2, 2).