We have, equation of the curve y(1+x2)=2−x...(i)
∴y⋅(0+2x)+(1+x2)⋅‌

dy

dx

=0−1‌ [on differentiating w.r.t.x] ‌
⇒2‌xy+(1+x2)‌

dy

dx

=−1
⇒‌

dy

dx

=‌

−1−2‌xy

1+x2

...‌ (ii) ‌
Since, the given curve passes through x− axis i.e., y=0.
∴O(1+x2)=2−x‌ [using Eq.(i)] ‌
⇒x=2
So, the curve passes through the point (2,0).
∴(‌

dy

dx

)(2,0)=‌

−1−2×0

1+22

=−‌

1

5

=‌ slope of the curve ‌
∴ slope of tangent to the curve =−‌

1

5

∴ Equation of tangent of the curve passing through (2,0) is
y−0=−‌

1

5

(x−2)
⇒5y=−x+2
⇒5y+x=2