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

The correct option is x + y = √2.

To find the equation of the tangent to a circle centered at the origin $(0,0)$ with radius $r=1$ at a given point $(x_1,y_1)$ on the circle, we can use the standard equation of a circle and its tangent property.

The general equation of a circle centered at the origin is:
$x^2+y^2=r^2$

For a radius $r=1$, the equation of the circle is:
$x^2+y^2=1$

The equation of the tangent to this circle at any point $(x_1,y_1)$ on the circle is given by:
$x{x}_1+y{y}_1=1$

Here, the point of tangency is:
$(x_1,y_1)=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$

Substituting these coordinates into the tangent equation, we get:
$x\left(\frac{1}{\sqrt{2}}\right)+y\left(\frac{1}{\sqrt{2}}\right)=1$

This simplifies to:
$\frac{x+y}{\sqrt{2}}=1$

Multiplying both sides by $\sqrt{2}$ yields the final equation of the tangent line:
$x+y=\sqrt{2}$