The straight line x + 2y = 1 meets the coordinate axes at A and B. A circle is drawn through A, B, and the origin. The sum of perpendicular distances from A and B on the tangent to the circle at the origin is:

Let the equation of the line be x + 2y = 1.
When x = 0, 2y = 1 => y = 1/2. So A = (0, 1/2).
When y = 0, x = 1. So B = (1, 0).
Let the equation of the circle passing through A(0, 1/2), B(1, 0), and O(0, 0) be x² + y² + 2gx + 2fy + c = 0.
Since the circle passes through (0, 0), c = 0.
Since the circle passes through (0, 1/2), (1/2)² + 2f(1/2) = 0 => 1/4 + f = 0 => f = -1/4.
Since the circle passes through (1, 0), 1 + 2g = 0 => g = -1/2.
Therefore, the equation of the circle is x² + y² - x - y/2 = 0.
The equation of the tangent at the origin is -x - y/2 = 0 => x + y/2 = 0 => 2x + y = 0.
The perpendicular distance from A(0, 1/2) to 2x + y = 0 is |2(0) + 1/2|/√(2² + 1²) = 1/(2√5) = √5/10.
The perpendicular distance from B(1, 0) to 2x + y = 0 is |2(1) + 0|/√(2² + 1²) = 2/√5 = 2√5/5.
Sum of perpendicular distances = √5/10 + 2√5/5 = 5√5/10 = √5/2.
Let's verify the equation of the circle using the general equation of a circle passing through three points (x1, y1), (x2, y2), (x3, y3).
The equation is given by:
|x² + y² x y 1|
|x1² + y1² x1 y1 1| = 0
|x2² + y2² x2 y2 1|
|x3² + y3² x3 y3 1|
For points (0, 0), (0, 1/2), (1, 0):
|x² + y² x y 1|
|0 0 0 1/2 1| = 0
|1 1 0 1 0|
|0 0 0 1 0|
This simplifies to x² + y² - x - y/2 = 0
The tangent at (0,0) is -x - y/2 = 0 => 2x+y = 0
Perpendicular distance from (0,1/2) to 2x+y=0 is |1/2|/√5 = 1/2√5
Perpendicular distance from (1,0) to 2x+y=0 is |2|/√5 = 2/√5
Sum = 1/2√5 + 2/√5 = 5/2√5 = √5/2 = 2√5/4 = √5/2
The sum of perpendicular distances from A and B on the tangent at the origin is √5/2 = √5/2 = √5/2 = √5/2 = √5/2 = √5/2 = √5/2 = 2.236/2 = 1.118 ≈ √5/2 ≈ 1.118
2√5/5 + √5/10 = 5√5/10 = √5/2