$Finding the Tangent Line to a Unit Circle at a Given Point$

Consider the unit circle centered at the origin with the equation:

$x^2 + y^2 = 1.$

To find the equation of the tangent line to the circle at a given point $P(a, b)$ on the circle, we follow these steps:

1. Verify that $P(a, b)$ lies on the circle, i.e., $a^2 + b^2 = 1$.

2. The slope of the radius to point $P(a, b)$ is $\frac{b}{a}.$

3. The slope of the tangent line at $P(a, b)$ is the negative reciprocal of the slope of the radius, which is $-\frac{a}{b}.$

4. Using the point-slope form of the line equation, we have:

$y – b = -\frac{a}{b}(x – a).$

Simplifying this, we get:

$bx + ay = 1.$

Thus, the equation of the tangent line to the unit circle at $P(a, b)$ is:

$bx + ay = 1.$

Given the unit circle with equation:

$x^2 + y^2 = 1,$

and a point $P(a, b)$ on the circle where $a^2 + b^2 = 1$. Follow these steps to find the tangent line:

1. Verify $a^2 + b^2 = 1$.

2. The slope of the radius to $P(a, b)$ is $m_r = frac{b}{a}.$

3. The tangent line slope is the negative reciprocal: $m_t = -frac{a}{b}.$

4. Using point-slope form:

$y – b = -frac{a}{b}(x – a).$

5. Simplifying, we get:

$bx + ay = 1.$

The tangent line equation is:

$bx + ay = 1.$

Given the unit circle $x^2 + y^2 = 1$ and a point $P(a, b)$ on it:

1. Confirm $a^2 + b^2 = 1.$

2. Radius slope to $P(a, b)$: $frac{b}{a}.$

3. Tangent slope: $-frac{a}{b}.$

4. Point-slope form: $y – b = -frac{a}{b}(x – a).$

5. Simplify: $bx + ay = 1.$

Equation of the tangent line: $bx + ay = 1.$

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