chord `AB` of the circle `x^2+y^2=100` passes through the point `(7,1) ` and subtends are angle of `60^@` at the circumference of the circle. if `m_1` and `m_2` are slopes of two such chords then the value of `m_1*m_2` is

Options

A. `-1`
B. 1
C. `7//12`
D. `-3`

Explanation

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

### Step 1: Understand the Circle and the Chord
The equation of the circle is given as \( x^2 + y^2 = 100 \). This means the radius \( r \) of the circle is \( \sqrt{100} = 10 \).

### Step 2: Equation of the Chord
The chord \( AB \) passes through the point \( (7, 1) \) and has a slope \( m \). The equation of the line (chord) can be written using the point-slope form:
\[ y - 1 = m(x - 7) \]
Rearranging gives:
\[ mx - y + (1 - 7m) = 0 \]

### Step 3: Perpendicular Distance from the Origin
The perpendicular distance \( d \) from the origin \( (0, 0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
Here, \( A = m \), \( B = -1 \), and \( C = 1 - 7m \). Thus, the distance from the origin to the chord is:
\[ d = \frac{|m \cdot 0 + (-1) \cdot 0 + (1 - 7m)|}{\sqrt{m^2 + 1}} = \frac{|1 - 7m|}{\sqrt{m^2 + 1}} \]
Since the chord subtends an angle of \( 60^\circ \) at the circumference, the perpendicular distance from the center of the circle to the chord is \( \frac{r}{2} = \frac{10}{2} = 5 \).

### Step 4: Set Up the Equation
Setting the perpendicular distance equal to 5, we have:
\[ \frac{|1 - 7m|}{\sqrt{m^2 + 1}} = 5 \]
Squaring both sides gives:
\[ (1 - 7m)^2 = 25(m^2 + 1) \]

### Step 5: Expand and Rearrange
Expanding both sides:
\[ 1 - 14m + 49m^2 = 25m^2 + 25 \]
Rearranging gives:
\[ 49m^2 - 25m^2 - 14m + 1 - 25 = 0 \]
This simplifies to:
\[ 24m^2 - 14m - 24 = 0 \]

### Step 6: Find the Product of the Slopes
Using the quadratic formula \( ax^2 + bx + c = 0 \), the product of the roots \( m_1 \cdot m_2 \) is given by \( \frac{c}{a} \):
\[ m_1 \cdot m_2 = \frac{-24}{24} = -1 \]

### Conclusion
Thus, the value of \( m_1 \cdot m_2 \) is \( -1 \).