Find the equations of the tangents to the unit circle at the point $left( frac{1}{sqrt{2}}, frac{1}{sqrt{2}}
ight)$

To find the equation of the tangent to the unit circle at a given point, we can use the formula for the tangent line to a circle:

$x_1 x + y_1 y = 1$

where $(x_1, y_1)$ is the point of tangency. Here, $x_1 = \frac{1}{\sqrt{2}}$ and $y_1 = \frac{1}{\sqrt{2}}$. Substituting these values into the formula, we get:

$\frac{1}{\sqrt{2}} x + \frac{1}{\sqrt{2}} y = 1$

Multiplying both sides by $\sqrt{2}$ to simplify, we obtain:

$x + y = \sqrt{2}$

Therefore, the equation of the tangent is:

$x + y = \sqrt{2}$

To find the tangent line to the unit circle at the point $left( frac{1}{sqrt{2}}, frac{1}{sqrt{2}}
ight)$, we use the fact that the slope of the radius at this point is the same as the slope of the line from the origin to this point. The slope is:

$ ext{slope} = frac{frac{1}{sqrt{2}} – 0}{frac{1}{sqrt{2}} – 0} = 1$

The slope of the tangent line is the negative reciprocal of this slope, which is:

$-1$

Using the point-slope form of the equation of a line, $y – y_1 = m(x – x_1)$, with $m = -1$ and $(x_1, y_1) = left( frac{1}{sqrt{2}}, frac{1}{sqrt{2}}
ight)$, we have:

$y – frac{1}{sqrt{2}} = -1 left( x – frac{1}{sqrt{2}}
ight)$

Simplifying, we get:

$y – frac{1}{sqrt{2}} = -x + frac{1}{sqrt{2}}$

Adding $ frac{1}{sqrt{2}} $ to both sides:

$y = -x + frac{2}{sqrt{2}}$

Simplifying further:

$y = -x + sqrt{2}$

Therefore, the equation of the tangent line is:

$x + y = sqrt{2}$

To find the tangent to the unit circle at $left( frac{1}{sqrt{2}}, frac{1}{sqrt{2}}
ight)$, use the tangent line formula:

$x_1 x + y_1 y = 1$

Substituting $x_1 = frac{1}{sqrt{2}}$ and $y_1 = frac{1}{sqrt{2}}$, we get:

$frac{1}{sqrt{2}} x + frac{1}{sqrt{2}} y = 1$

Simplifying:

$x + y = sqrt{2}$

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