Question:

Step 1: Determine the coordinates of points A and B and the equation of the circle. The given line is $x + 2y = 1$. To find point A, where the line meets the y-axis, set $x=0$: $0 + 2y = 1 \implies y = 1/2$. Thus, $A = (0, 1/2)$. To find point B, where the line meets the x-axis, set $y=0$: $x + 2(0) = 1 \implies x = 1$. Thus, $B = (1, 0)$. The circle passes through the origin $O(0,0)$, $A(0, 1/2)$, and $B(1, 0)$. The general equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$. The perpendicular bisector of the chord OA (on the y-axis) passes through its midpoint $(0, 1/4)$ and is horizontal, so its equation is $y = 1/4$. The perpendicular bisector of the chord OB (on the x-axis) passes through its midpoint $(1/2, 0)$ and is vertical, so its equation is $x = 1/2$. The center of the circle, C, is the intersection of these bisectors: $C = (1/2, 1/4)$. The radius $R$ of the circle is the distance from the center C to any of the points O, A, or B. Using O(0,0): $R = \sqrt{(1/2 - 0)^2 + (1/4 - 0)^2} = \sqrt{(1/2)^2 + (1/4)^2} = \sqrt{1/4 + 1/16} = \sqrt{4/16 + 1/16} = \sqrt{5/16} = \frac{\sqrt{5}}{4}$. Step 2: Find the equation of the tangent to the circle at the origin O(0,0). The slope of the radius CO is $m_{CO} = \frac{1/4 - 0}{1/2 - 0} = \frac{1/4}{1/2} = 1/2$. The tangent at O is perpendicular to CO, so its slope is $m_{tan,O} = -1/m_{CO} = -2$. The equation of the tangent at O passing through $(0,0)$ is: $y - 0 = -2(x - 0) \implies y = -2x \implies 2x + y = 0$. Step 3: Find the equation of the tangent to the circle at B(1,0). The slope of the radius CB is $m_{CB} = \frac{0 - 1/4}{1 - 1/2} = \frac{-1/4}{1/2} = -1/2$. The tangent at B is perpendicular to CB, so its slope is $m_{tan,B} = -1/m_{CB} = 2$. The equation of the tangent at B passing through $(1,0)$ is: $y - 0 = 2(x - 1) \implies y = 2x - 2 \implies 2x - y - 2 = 0$. Step 4: Calculate the sum of distances. The distance from the center of a circle to any tangent line is equal to the radius of the circle. The distance $l_1$ from the center $C(1/2, 1/4)$ to the tangent at O ($2x+y=0$) is: $l_1 = \frac{|2(1/2) + 1/4|}{\sqrt{2^2 + 1^2}} = \frac{|1 + 1/4|}{\sqrt{5}} = \frac{5/4}{\sqrt{5}} = \frac{5}{4\sqrt{5}} = \frac{\sqrt{5}}{4}$. The distance $l_2$ from the center $C(1/2, 1/4)$ to the tangent at B ($2x-y-2=0$) is: $l_2 = \frac{|2(1/2) - 1/4 - 2|}{\sqrt{2^2 + (-1)^2}} = \frac{|1 - 1/4 - 2|}{\sqrt{5}} = \frac{|-5/4|}{\sqrt{5}} = \frac{5/4}{\sqrt{5}} = \frac{\sqrt{5}}{4}$. The sum of these distances is: $l_1 + l_2 = \frac{\sqrt{5}}{4} + \frac{\sqrt{5}}{4} = \frac{2\sqrt{5}}{4} = \frac{\sqrt{5}}{2}$. Numerically, $\sqrt{5} \approx 2.236067977$. So, $l_1 + l_2 = \frac{2.236067977}{2} \approx 1.118033988$. The sum of the distances is approximately $1.1180$.