A line `2x + y = 1` intersect co-ordinate axis at points `A` and `B`. A circle is drawn passing through origin and point `A` & `B`. If perpendicular from point `A` and `B` are drawn on tangent to the circle at origin then sum of perpendicular distance is (A) `5/sqrt2` (B) `sqrt5/2` (C) `sqrt5/4` (D) `5/2`

Options

A. `(sqrt(5))/(4)`
B. `(sqrt(5))/(2)`
C. `2sqrt(5)`
D. `24sqrt(5)`

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Find the points of intersection A and B
The line given is \(2x + y = 1\).

**Finding point A (x-intercept)**:
To find the x-intercept, set \(y = 0\):
\[ 2x + 0 = 1 \implies x = \frac{1}{2} \]
Thus, point A is \(\left(\frac{1}{2}, 0\right)\).

**Finding point B (y-intercept)**:
To find the y-intercept, set \(x = 0\):
\[ 2(0) + y = 1 \implies y = 1 \]
Thus, point B is \((0, 1)\).

### Step 2: Determine the equation of the circle
The circle passes through the origin \(O(0, 0)\), point A \(\left(\frac{1}{2}, 0\right)\), and point B \((0, 1)\). The general equation of a circle can be expressed as:
\[ (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \]
Using points \(O(0, 0)\), A \(\left(\frac{1}{2}, 0\right)\), and B \((0, 1)\), we can write the equation of the circle as:
\[ x(x - \frac{1}{2}) + y(y - 1) = 0 \]
Expanding this gives:
\[ x^2 - \frac{1}{2}x + y^2 - y = 0 \]

### Step 3: Find the tangent at the origin
The tangent to the circle at the origin can be found using the derivative or the general form of the circle. The tangent line at the origin is given by:
\[ x + 2y = 0 \]
This can be rearranged to:
\[ x + 2y = 0 \]

### Step 4: Calculate the perpendicular distances from points A and B to the tangent
**Perpendicular distance from point A \(\left(\frac{1}{2}, 0\right)\)**:
Using the formula for the distance from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\):
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For the line \(x + 2y = 0\) (where \(A = 1\), \(B = 2\), \(C = 0\)):
\[ d_A = \frac{|1 \cdot \frac{1}{2} + 2 \cdot 0 + 0|}{\sqrt{1^2 + 2^2}} = \frac{\frac{1}{2}}{\sqrt{5}} = \frac{1}{2\sqrt{5}} \]

**Perpendicular distance from point B \((0, 1)\)**:
\[ d_B = \frac{|1 \cdot 0 + 2 \cdot 1 + 0|}{\sqrt{1^2 + 2^2}} = \frac{2}{\sqrt{5}} \]

### Step 5: Sum of the perpendicular distances
Now, we sum the distances:
\[ d_A + d_B = \frac{1}{2\sqrt{5}} + \frac{2}{\sqrt{5}} = \frac{1 + 4}{2\sqrt{5}} = \frac{5}{2\sqrt{5}} = \frac{\sqrt{5}}{2} \]

### Final Answer
Thus, the sum of the perpendicular distances is \(\frac{\sqrt{5}}{2}\).