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(A) The line $x + 2y = 1$ intersects the $x$-axis at $A(1, 0)$ and the $y$-axis at $B(0, 1/2)$.Since the circle passes through the origin $(0, 0)$,$A(

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$\frac{\sqrt{5}}{2}$

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(A) The line $x + 2y = 1$ intersects the $x$-axis at $A(1, 0)$ and the $y$-axis at $B(0, 1/2)$.Since the circle passes through the origin $(0, 0)$,$A(1, 0)$,and $B(0, 1/2)$,and the triangle $OAB$ is a right-angled triangle at the origin,the segment $AB$ is the diameter of the circle.The equation of the circle with diameter $AB$ is $(x - 0)(x - 1) + (y - 0)(y - 1/2) = 0$,which simplifies to $x^2 + y^2 - x - \frac{1}{2}y = 0$.The tangent to the circle at the origin $(0, 0)$ is found by replacing $x^2$ with $x \cdot 0$,$y^2$ with $y \cdot 0$,$x$ with $\frac{x+0}{2}$,and $y$ with $\frac{y+0}{2}$.This gives $0 + 0 - \frac{x}{2} - \frac{1}{2} \cdot \frac{y}{2} = 0$,which simplifies to $-\frac{x}{2} - \frac{y}{4} = 0$,or $2x + y = 0$.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 $B(0, 1/2)$ to the line $2x + y = 0$ is $d_2 = \frac{|2(0) + 1/2|}{\sqrt{2^2 + 1^2}} = \frac{1/2}{\sqrt{5}} = \frac{1}{2\sqrt{5}}$.The sum of the distances is $d_1 + d_2 = \frac{2}{\sqrt{5}} + \frac{1}{2\sqrt{5}} = \frac{4+1}{2\sqrt{5}} = \frac{5}{2\sqrt{5}} = \frac{\sqrt{5}}{2}$.

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

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