The tangent to the circle centred at (0, 0) and with radius = 1 at point $\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$ on it is given by

$x+y=\sqrt{2}$

centre (0, 0) radius = 1

eqn of circle: $(x-0)^2+(y-0)^=1^2$

$⇒x^2+y^2=1$

finding slope of tangent at $\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$

differentiating eqn w.r.t x

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

$\frac{2}{\sqrt{2}}+\frac{2}{\sqrt{2}}\frac{dy}{dx}]_{(x,y)=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)}=0$

so $\frac{dy}{dx}=-1$ (slope)

eq: of tangent → $(y-\frac{1}{\sqrt{2}})=-1(x-\frac{1}{\sqrt{2}})$

$⇒x+y=\sqrt{2}$