Key Concepts and Formulas

• Intercept Form of a Line: The equation of a line in intercept form is given by $\frac{x}{a} + \frac{y}{b} = 1$, where $a$ and $b$ are the x and y intercepts, respectively.
• Equation of a Circle Passing Through the Origin and Axial Intercepts: If a circle passes through the origin and intercepts the x-axis at $(a, 0)$ and the y-axis at $(0, b)$, its equation is $x^2 + y^2 - ax - by = 0$.
• Equation of Tangent to a Circle at the Origin: For a circle with the equation $x^2 + y^2 + 2gx + 2fy = 0$, the equation of the tangent at the origin is $gx + fy = 0$.
• Perpendicular Distance from a Point to a Line: The perpendicular distance from a point $(x_1, y_1)$ to a line $ax + by + c = 0$ is given by $\frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$.

Step-by-Step Solution

Step 1: Find the coordinates of A and B

The line $x + 2y = 1$ intersects the x-axis when $y = 0$, so $x = 1$. Thus, $A = (1, 0)$. The line intersects the y-axis when $x = 0$, so $2y = 1$, which means $y = \frac{1}{2}$. Thus, $B = (0, \frac{1}{2})$.

Step 2: Find the equation of the circle

Since the circle passes through the origin, A(1,0) and B(0, 1/2), its equation is of the form $x^2 + y^2 - x - \frac{1}{2}y = 0$.

Step 3: Find the equation of the tangent to the circle at the origin

Comparing the circle's equation, $x^2 + y^2 - x - \frac{1}{2}y = 0$, with the general form $x^2 + y^2 + 2gx + 2fy = 0$, we have $2g = -1$ and $2f = -\frac{1}{2}$, so $g = -\frac{1}{2}$ and $f = -\frac{1}{4}$. The equation of the tangent at the origin is $gx + fy = 0$, which gives $-\frac{1}{2}x - \frac{1}{4}y = 0$. Multiplying by -4, we get $2x + y = 0$.

Step 4: Find the perpendicular distance from A and B to the tangent

The perpendicular distance from $A(1, 0)$ to the line $2x + y = 0$ is $d_1 = \frac{|2(1) + 0|}{\sqrt{2^2 + 1^2}} = \frac{2}{\sqrt{5}}$.

The perpendicular distance from $B(0, \frac{1}{2})$ to the line $2x + y = 0$ is $d_2 = \frac{|2(0) + \frac{1}{2}|}{\sqrt{2^2 + 1^2}} = \frac{\frac{1}{2}}{\sqrt{5}} = \frac{1}{2\sqrt{5}}$.

Step 5: Calculate the sum of the perpendicular distances

The sum of the perpendicular distances is $d_1 + d_2 = \frac{2}{\sqrt{5}} + \frac{1}{2\sqrt{5}} = \frac{4 + 1}{2\sqrt{5}} = \frac{5}{2\sqrt{5}} = \frac{\sqrt{5}}{2}$.

Common Mistakes & Tips

• Be careful when identifying the coefficients $g$ and $f$ from the circle's equation to determine the tangent at the origin.
• Remember the formula for the perpendicular distance from a point to a line.
• Rationalize the denominator when simplifying expressions involving square roots.

Summary

We first found the coordinates of the points A and B where the line intersects the axes. Then, we determined the equation of the circle passing through A, B, and the origin. Next, we found the equation of the tangent to the circle at the origin. Finally, we calculated the perpendicular distances from A and B to the tangent and summed them to get the final answer.

Final Answer

The final answer is $\boxed{\frac{\sqrt{5}}{2}}$, which corresponds to option (B).