We have, equation of the curve $y (1+x^2) =2-x \dots (i)$$\therefore y \cdot (0+2x) + (1+x^2) \cdot \frac{dy}{dx} = 0-1 \; \text{ [on differentiating w.r.t.x] }\;$$\Rightarrow 2xy + (1+x^2) \; \frac{dy}{dx} = -1$$\Rightarrow \; \frac{dy}{dx} = \; \frac{-1-2xy}{1+x^2} \dots \; \text{ (ii) }\;$Since, the given curve passes through $x-$ axis i.e., $y=0$.$\therefore O (1+x^2) = 2-x \; \text{ [using Eq.(i)] }\;$$\Rightarrow x=2$So, the curve passes through the point $(2,0)$.$\therefore \left(\frac{dy}{dx}\right)_{(2,0)} = \frac{-1-2 \times 0}{1+2^2} = -\frac{1}{5} = \; \text{ slope of the curve }\;$$\therefore$ slope of tangent to the curve $= -\; \frac{1}{5}$$\therefore$ Equation of tangent of the curve passing through $(2,0)$ is$y-0 = -\; \frac{1}{5}(x-2)$$\Rightarrow 5y = -x + 2$$\Rightarrow 5y + x = 2$