A line makes equal intercepts of length a on the coordinate axes. A circle is circumscribed about the triangle which the line makes with the coordinate axes. The sum of the distance of the vertices of the triangle from the tangent to this circle at the origin is

Options

A. `a//sqrt2`
B. `a`
C. `asqrt2`
D. `a + sqrt2`

Answer

To solve the problem step by step, we need to find the sum of the distances of the vertices of the triangle formed by the line making equal intercepts on the coordinate axes from the tangent to the circumscribed circle at the origin.

### Step 1: Determine the equation of the line
The line makes equal intercepts of length \( a \) on the coordinate axes. Therefore, the intercepts on the x-axis and y-axis are both \( a \). The equation of the line can be expressed as:
\[ x + y = a \]

### Step 2: Identify the vertices of the triangle
The line intersects the x-axis at the point \( (a, 0) \) and the y-axis at the point \( (0, a) \). The third vertex of the triangle is the origin \( (0, 0) \). Thus, the vertices of the triangle are:
- \( A(a, 0) \)
- \( B(0, a) \)
- \( O(0, 0) \)

### Step 3: Find the circumcircle of the triangle
The circumcircle of a triangle can be defined by its circumradius \( R \) and its circumcenter. For triangle \( AOB \):
- The midpoint of the line segment \( AB \) is given by:
\[ \left( \frac{a}{2}, \frac{a}{2} \right) \]
- The radius \( R \) of the circumcircle can be calculated using the distance from the circumcenter to any vertex. The distance from the origin \( O(0, 0) \) to the midpoint \( \left( \frac{a}{2}, \frac{a}{2} \right) \) is:
\[ R = \sqrt{\left( \frac{a}{2} \right)^2 + \left( \frac{a}{2} \right)^2} = \sqrt{\frac{a^2}{4} + \frac{a^2}{4}} = \sqrt{\frac{a^2}{2}} = \frac{a}{\sqrt{2}} \]

### Step 4: Write the equation of the circumcircle
The equation of the circumcircle with center at the origin \( (0, 0) \) and radius \( R \) is:
\[ x^2 + y^2 = \left(\frac{a}{\sqrt{2}}\right)^2 = \frac{a^2}{2} \]

### Step 5: Find the equation of the tangent at the origin
The tangent to the circle at the origin can be expressed as:
\[ 0 = 0 \cdot x + 0 \cdot y - \frac{a^2}{2} \]
This simplifies to:
\[ x + y = \frac{a}{\sqrt{2}} \]

### Step 6: Calculate the distances from the vertices to the tangent line
Using the formula for the distance from a point \( (x_1, y_1) \) to the line \( Ax + By + C = 0 \):
\[ \text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
For the line \( x + y - \frac{a}{\sqrt{2}} = 0 \) (where \( A = 1, B = 1, C = -\frac{a}{\sqrt{2}} \)):
1. Distance from \( A(a, 0) \):
\[ \text{Distance}_A = \frac{|1 \cdot a + 1 \cdot 0 - \frac{a}{\sqrt{2}}|}{\sqrt{1^2 + 1^2}} = \frac{|a - \frac{a}{\sqrt{2}}|}{\sqrt{2}} = \frac{a(\sqrt{2} - 1)}{\sqrt{2}} \]

2. Distance from \( B(0, a) \):
\[ \text{Distance}_B = \frac{|1 \cdot 0 + 1 \cdot a - \frac{a}{\sqrt{2}}|}{\sqrt{2}} = \frac{|a - \frac{a}{\sqrt{2}}|}{\sqrt{2}} = \frac{a(\sqrt{2} - 1)}{\sqrt{2}} \]

3. Distance from \( O(0, 0) \):
\[ \text{Distance}_O = \frac{|1 \cdot 0 + 1 \cdot 0 - \frac{a}{\sqrt{2}}|}{\sqrt{2}} = \frac{|\frac{a}{\sqrt{2}}|}{\sqrt{2}} = \frac{a}{2} \]

### Step 7: Sum the distances
Now, we sum the distances:
\[ \text{Total Distance} = \text{Distance}_A + \text{Distance}_B + \text{Distance}_O = \frac{a(\sqrt{2} - 1)}{\sqrt{2}} + \frac{a(\sqrt{2} - 1)}{\sqrt{2}} + \frac{a}{2} \]
Simplifying this gives:
\[ \text{Total Distance} = 2 \cdot \frac{a(\sqrt{2} - 1)}{\sqrt{2}} + \frac{a}{2} = \frac{2a(\sqrt{2} - 1) + a\sqrt{2}}{2} = \frac{a(2\sqrt{2} - 2 + \sqrt{2})}{2} = \frac{a(3\sqrt{2} - 2)}{2} \]

### Conclusion
Thus, the sum of the distances of the vertices of the triangle from the tangent to the circle at the origin is:
\[ \frac{a(3\sqrt{2} - 2)}{2} \]