A circle `(x-2)^2+(y-3)^2=16` is given for which two lines `L_1:2x+3y=7` and `L_2:3x+2y=11` act as chords . If infinitely more chords are drawn parallel to `L_1` and `L_2`. A curve is drawn by joining mid points of chords parallel to `L_1` and another curve is drawn by joining mid points of chords parallel to `L_2`.
(Both the curves are then extended on both sides) , then the acute angle between the curves at the point of their intersection is

Options

A. `pi/2`
B. `tan^(-1)(5/12)`
C. `tan^(-1)(7/12)`
D. `pi/6`

Answer

To find the acute angle between the curves formed by the midpoints of the chords parallel to the lines \( L_1 \) and \( L_2 \) at their intersection point, we can follow these steps:

### Step 1: Identify the Circle and Lines
The given circle is:
\[ (x-2)^2 + (y-3)^2 = 16 \]
This indicates that the center of the circle is at \( (2, 3) \) and the radius is \( 4 \).

The lines are:
\[ L_1: 2x + 3y = 7 \]
\[ L_2: 3x + 2y = 11 \]

### Step 2: Find the Slopes of the Lines
To find the slopes of the lines, we can rewrite them in slope-intercept form \( y = mx + b \).

For \( L_1 \):
\[ 3y = 7 - 2x \implies y = -\frac{2}{3}x + \frac{7}{3} \]
Thus, the slope \( m_1 = -\frac{2}{3} \).

For \( L_2 \):
\[ 2y = 11 - 3x \implies y = -\frac{3}{2}x + \frac{11}{2} \]
Thus, the slope \( m_2 = -\frac{3}{2} \).

### Step 3: Find the Slopes of the Perpendicular Lines
The slopes of the lines that are perpendicular to \( L_1 \) and \( L_2 \) can be calculated as follows:
- The slope \( m_1' \) perpendicular to \( L_1 \) is given by:
\[ m_1' = \frac{3}{2} \]
- The slope \( m_2' \) perpendicular to \( L_2 \) is given by:
\[ m_2' = \frac{2}{3} \]

### Step 4: Use the Formula for the Angle Between Two Lines
The formula for the angle \( \theta \) between two lines with slopes \( m_1' \) and \( m_2' \) is:
\[ \tan \theta = \left| \frac{m_1' - m_2'}{1 + m_1' m_2'} \right| \]

Substituting the values:
\[ \tan \theta = \left| \frac{\frac{3}{2} - \frac{2}{3}}{1 + \left(\frac{3}{2} \cdot \frac{2}{3}\right)} \right| \]

### Step 5: Simplify the Expression
Calculating the numerator:
\[ \frac{3}{2} - \frac{2}{3} = \frac{9}{6} - \frac{4}{6} = \frac{5}{6} \]

Calculating the denominator:
\[ 1 + \left(\frac{3}{2} \cdot \frac{2}{3}\right) = 1 + 1 = 2 \]

Thus, we have:
\[ \tan \theta = \left| \frac{\frac{5}{6}}{2} \right| = \frac{5}{12} \]

### Step 6: Find the Angle
To find \( \theta \):
\[ \theta = \tan^{-1}\left(\frac{5}{12}\right) \]

This angle \( \theta \) is the acute angle between the curves at the point of their intersection.

### Final Answer
The acute angle between the curves at the point of their intersection is:
\[ \theta = \tan^{-1}\left(\frac{5}{12}\right) \]

---