If the circle `C_(1)` touches x-axis and the line `y=xtantheta(tanthetagt0)` in first quadrant and circle`C_(2)` touches the `y=xtantheta` at the same point at which `C_(1)` touches it such that ratio of radius of `C_(1)` and `C_(2)` is 2:1, then `tan(theta)/(2)=sqrt(a-B)/(c)` where a,b,c,epsilon N and `HCF(b,c)=1`

Options

A. `a=13`
B. `b=3`
C. `c=2`
D. `a=17`

Explanation

To solve the problem step by step, let's break it down:

### Step 1: Understand the Geometry
We have two circles, \( C_1 \) and \( C_2 \). Circle \( C_1 \) touches the x-axis and the line \( y = x \tan \theta \) in the first quadrant. Circle \( C_2 \) touches the same line at the same point as \( C_1 \), and the ratio of their radii is \( 2:1 \).

### Step 2: Define the Radii and Centers
Let the radius of circle \( C_1 \) be \( r_1 \) and the radius of circle \( C_2 \) be \( r_2 \). Given the ratio \( \frac{r_1}{r_2} = 2 \), we can express \( r_2 \) as:
\[ r_2 = \frac{r_1}{2} \]

### Step 3: Determine the Touching Point
Let the point where both circles touch the line \( y = x \tan \theta \) be \( P(d, d \tan \theta) \). The distance from the center of \( C_1 \) to the x-axis is \( r_1 \), and the center of \( C_1 \) can be denoted as \( O_1(d, r_1) \).

### Step 4: Use Right Triangle Properties
In triangle \( O_1 P \):
\[ \tan \left(\frac{\theta}{2}\right) = \frac{r_1}{d} \]
Thus,
\[ d = r_1 \cot \left(\frac{\theta}{2}\right) \]

### Step 5: Analyze Circle \( C_2 \)
For circle \( C_2 \), its center \( O_2 \) can be denoted as \( (d, r_2) \). The distance from \( O_2 \) to the line \( y = x \tan \theta \) gives:
\[ \tan \left(90^\circ - \frac{\theta}{2}\right) = \frac{r_2}{d} \]
Thus,
\[ \tan \left(45^\circ - \frac{\theta}{2}\right) = \frac{r_2}{d} \]

### Step 6: Substitute for \( d \)
Substituting \( d \) from the previous step:
\[ \tan \left(45^\circ - \frac{\theta}{2}\right) = \frac{r_2}{r_1 \cot \left(\frac{\theta}{2}\right)} \]

### Step 7: Simplify the Equation
Using the identity \( \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \):
\[ \tan \left(45^\circ - \frac{\theta}{2}\right) = \frac{1 - \tan \left(\frac{\theta}{2}\right)}{1 + \tan \left(\frac{\theta}{2}\right)} \]
Substituting this into our equation gives:
\[ \frac{1 - \tan \left(\frac{\theta}{2}\right)}{1 + \tan \left(\frac{\theta}{2}\right)} = \frac{r_2}{r_1 \cot \left(\frac{\theta}{2}\right)} \]

### Step 8: Substitute \( r_2 \)
Substituting \( r_2 = \frac{r_1}{2} \):
\[ \frac{1 - \tan \left(\frac{\theta}{2}\right)}{1 + \tan \left(\frac{\theta}{2}\right)} = \frac{\frac{r_1}{2}}{r_1 \cot \left(\frac{\theta}{2}\right)} \]
This simplifies to:
\[ \frac{1 - \tan \left(\frac{\theta}{2}\right)}{1 + \tan \left(\frac{\theta}{2}\right)} = \frac{1}{2 \cot \left(\frac{\theta}{2}\right)} \]

### Step 9: Solve for \( \tan \left(\frac{\theta}{2}\right) \)
Cross-multiplying and simplifying leads to a quadratic equation in terms of \( \tan \left(\frac{\theta}{2}\right) \):
\[ \tan^2 \left(\frac{\theta}{2}\right) + 3 \tan \left(\frac{\theta}{2}\right) - 2 = 0 \]

### Step 10: Use the Quadratic Formula
Using the quadratic formula:
\[ \tan \left(\frac{\theta}{2}\right) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 3, c = -2 \):
\[ \tan \left(\frac{\theta}{2}\right) = \frac{-3 \pm \sqrt{9 + 8}}{2} = \frac{-3 \pm \sqrt{17}}{2} \]

### Step 11: Choose the Positive Root
Since \( \tan \left(\frac{\theta}{2}\right) > 0 \):
\[ \tan \left(\frac{\theta}{2}\right) = \frac{-3 + \sqrt{17}}{2} \]

### Step 12: Express in Required Form
We need to express \( \frac{\tan \theta}{2} \) in the form \( \sqrt{a - b}/c \):
\[ \frac{\tan \theta}{2} = \frac{\sqrt{17} - 3}{2} \]
Thus, \( a = 17, b = 3, c = 2 \).

### Final Answer
The values are \( a = 17, b = 3, c = 2 \) and \( \text{HCF}(b, c) = 1 \).