Find the equations of straight lines which pass through the intersection of the lines `x -2y-5=0, 7x +y =50` & divide the circumference of the circle `x^2 + y^2 =100` into two arcs whose lengths are in the ratio 2:1.

Options

A. `3x+4y-25=0`
B. `4x-3y-25=0`
C. `3x+2y-23=0`
D. `2x-3y-11=0`

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Find the intersection of the lines
We have two lines given by the equations:

1. \( x - 2y - 5 = 0 \) (Equation 1)
2. \( 7x + y - 50 = 0 \) (Equation 2)

To find the intersection point, we can solve these equations simultaneously.

From Equation 1, we can express \( x \) in terms of \( y \):
\[ x = 2y + 5 \]

Now, substitute this value of \( x \) into Equation 2:
\[ 7(2y + 5) + y - 50 = 0 \]
\[ 14y + 35 + y - 50 = 0 \]
\[ 15y - 15 = 0 \]
\[ y = 1 \]

Now substitute \( y = 1 \) back into the expression for \( x \):
\[ x = 2(1) + 5 = 7 \]

Thus, the intersection point is \( (7, 1) \).

### Step 2: Determine the angle subtended by the arcs
The circumference of the circle \( x^2 + y^2 = 100 \) has a radius \( r = 10 \). The total angle around a circle is \( 360^\circ \). Since the lengths of the arcs are in the ratio \( 2:1 \), the angle subtended by the chord \( AB \) at the center \( O \) will be:
\[ \text{Total angle} = 360^\circ \]
\[ \text{Angle subtended by arc 1} = \frac{2}{3} \times 360^\circ = 240^\circ \]
\[ \text{Angle subtended by arc 2} = \frac{1}{3} \times 360^\circ = 120^\circ \]

### Step 3: Calculate the perpendicular distance from the center to the line
Let \( O \) be the center of the circle, which is at the origin \( (0, 0) \). The distance \( OM \) from the center to the chord \( AB \) can be calculated using the formula:
\[ OM = r \cos\left(\frac{\theta}{2}\right) \]
where \( \theta = 120^\circ \):
\[ OM = 10 \cos(60^\circ) = 10 \cdot \frac{1}{2} = 5 \]

### Step 4: Find the equation of the line through the intersection point
The equation of the line in point-slope form through the point \( (7, 1) \) is:
\[ y - 1 = m(x - 7) \]
Rearranging gives:
\[ mx - 7m - y + 1 = 0 \quad \Rightarrow \quad mx - y - 7m + 1 = 0 \]

### Step 5: Use the distance formula to find \( m \)
The perpendicular distance from the center \( O(0, 0) \) to the line is given by:
\[ \frac{|0 - 0 - 7m + 1|}{\sqrt{m^2 + 1}} = 5 \]
This simplifies to:
\[ \frac{|1 - 7m|}{\sqrt{m^2 + 1}} = 5 \]

Squaring both sides:
\[ (1 - 7m)^2 = 25(m^2 + 1) \]
Expanding and rearranging gives:
\[ 1 - 14m + 49m^2 = 25m^2 + 25 \]
\[ 24m^2 - 14m - 24 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula:
\[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{14 \pm \sqrt{(-14)^2 - 4 \cdot 24 \cdot (-24)}}{2 \cdot 24} \]
Calculating the discriminant:
\[ = \frac{14 \pm \sqrt{196 + 2304}}{48} = \frac{14 \pm \sqrt{2500}}{48} = \frac{14 \pm 50}{48} \]
This gives us two slopes:
\[ m_1 = \frac{64}{48} = \frac{4}{3}, \quad m_2 = \frac{-36}{48} = -\frac{3}{4} \]

### Step 7: Write the equations of the lines
Substituting back into the line equation:
1. For \( m = \frac{4}{3} \):
\[ 4x - 3y - 25 = 0 \]
2. For \( m = -\frac{3}{4} \):
\[ 3x + 4y - 25 = 0 \]

### Final Answer
The equations of the straight lines are:
1. \( 4x - 3y - 25 = 0 \)
2. \( 3x + 4y - 25 = 0 \)