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Step-by-step Solution
Step 1: Identify the intercepts of the line on the coordinate axes
The given line is
$ x + 2y = 1 $.
• At the x-intercept (point A), set $ y=0 $. Then $ x = 1 $, so $ A(1,0) $.
• At the y-intercept (point B), set $ x=0 $. Then $ 2y = 1 \implies y = \tfrac12 $, so $ B\bigl(0,\tfrac12\bigr) $.
Step 2: Find the equation of the circle passing through O, A, and B
Let the circle be
$ x^2 + y^2 + 2gx + 2fy = 0 $.
Since it passes through the origin $O(0,0)$, plugging $ x=0,\, y=0 $ gives
$ 0 + 0 + 0 + 0 = 0 $,
so there is no constant term.
1) Passing through $ A(1,0) $:
$ 1^2 + 0 + 2g(1) + 2f(0) = 0 \implies 1 + 2g = 0 \implies g = -\tfrac12. $
2) Passing through $ B\bigl(0,\tfrac12\bigr) $:
$ 0 + \bigl(\tfrac12\bigr)^2 + 2\bigl(-\tfrac12\bigr)(0) + 2f\bigl(\tfrac12\bigr) = 0
\implies \tfrac14 + f = 0 \implies f = -\tfrac14. $
Thus, the circle is
$ x^2 + y^2 - x - \tfrac12y = 0 $.
Step 3: Determine the tangent to the circle at the origin
1) The center of the circle is
$ (-g,\,-f) = \bigl(\tfrac12,\tfrac14\bigr). $
2) The line from the center $ C\bigl(\tfrac12,\tfrac14\bigr) $ to $ O(0,0) $ has slope
$ \frac{\tfrac14 - 0}{\tfrac12 - 0} = \tfrac12 $.
3) The tangent at $O$ is perpendicular to $CO$, so its slope is the negative reciprocal of $ \tfrac12 $, which is $ -2 $.
Hence the tangent line through the origin has equation
$ y = -2x $
or
$ 2x + y = 0 $.
Step 4: Calculate the perpendicular distances from A and B to this tangent
The perpendicular distance from a point $(x_1,y_1)$ to the line
$ Ax + By + C = 0 $
is
$ \dfrac{\bigl|Ax_1 + By_1 + C\bigr|}{\sqrt{A^2 + B^2}} $.
For the line $ 2x + y = 0 $:
$ A = 2,\; B = 1,\; C = 0,\; \sqrt{A^2 + B^2} = \sqrt{4+1} = \sqrt{5}. $
1) Distance from $ A(1,0) $:
$ d_A = \frac{|2\cdot 1 + 1 \cdot 0 + 0|}{\sqrt{5}} = \frac{2}{\sqrt{5}}. $
2) Distance from $ B\bigl(0,\tfrac12\bigr) $:
$ d_B = \frac{|2\cdot 0 + 1 \cdot \tfrac12 + 0|}{\sqrt{5}} = \frac{\tfrac12}{\sqrt{5}} = \frac{1}{2\sqrt{5}}. $
Step 5: Sum of the perpendicular distances
$ d_A + d_B
= \frac{2}{\sqrt{5}} + \frac{1}{2 \sqrt{5}}
= \frac{4}{2\sqrt{5}} + \frac{1}{2\sqrt{5}}
= \frac{5}{2\sqrt{5}}
= \frac{\sqrt{5}}{2}.
$
Thus, the required sum of perpendicular distances from $A$ and $B$ on the tangent to the circle at the origin is
$ \frac{\sqrt{5}}{2}.
$
