Determine the coordinates on the unit circle where the tangent line passes through the point $(1, 2)$

To find the points on the unit circle where the tangent line passes through the point $(1, 2)$, we can use the following steps:

1. The equation of the unit circle is given by $x^2 + y^2 = 1$.

2. The slope of the tangent line at any point $(a, b)$ on the circle is $-a/b$.

3. The equation of the tangent line at $(a, b)$ can be written as:

$y – b = -\frac{a}{b}(x – a)$

4. Substitute the point $(1, 2)$ into this equation to find $(a, b)$:

$2 – b = -\frac{a}{b}(1 – a)$

5. Simplify and solve for $(a, b)$.

After solving, we get:

$a = \frac{\sqrt{2}}{3}$ and $b = \frac{\sqrt{2}}{3}$.

Thus, the coordinates are: $(\frac{\sqrt{2}}{3}, \frac{\sqrt{2}}{3})$ and $(\frac{-\sqrt{2}}{3}, \frac{-\sqrt{2}}{3})$.

To find the coordinates on the unit circle where the tangent line passes through $(1, 2)$:

1. The unit circle equation is $x^2 + y^2 = 1$.

2. Find the slope of the tangent line at $(a, b)$: $-a/b$.

3. Use the point-slope form of the tangent line through $(1, 2)$:

$2 – b = -frac{a}{b}(1 – a)$.

Solving for $(a, b)$ gives:

$a = frac{sqrt{2}}{3}$ and $b = frac{sqrt{2}}{3}$.

Thus, the coordinates are: $(frac{sqrt{2}}{3}, frac{sqrt{2}}{3})$ and $(frac{-sqrt{2}}{3}, frac{-sqrt{2}}{3})$.

To find the tangent line points on the unit circle passing through $(1, 2)$:

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

2. Slope at $(a, b)$: $-a/b$.

3. Tangent line: $2 – b = -frac{a}{b}(1 – a)$.

Solutions: $(frac{sqrt{2}}{3}, frac{sqrt{2}}{3})$ and $(frac{-sqrt{2}}{3}, frac{-sqrt{2}}{3})$.

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