A straight line `l_(1)` with equation `x-2y+10=0` meets the circle with equation `x^(2)+y^(2)=100` at B in the first quadrant. A line through B, perpendiclar to `l_(1)` cuts the x - axis and y - axis at P and Q respectively. The area (in sq. units) of the triangle OPQ is (where, O is the origin)

Options

A. 120
B. 150
C. 100
D. 125

Answer

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

### Step 1: Find the intersection point B of the line and the circle.
The given line is \( l_1: x - 2y + 10 = 0 \) and the circle is \( x^2 + y^2 = 100 \).

First, we can express \( y \) in terms of \( x \) from the line equation:
\[ x - 2y + 10 = 0 \implies 2y = x + 10 \implies y = \frac{x + 10}{2} \]

Now, substitute \( y \) into the circle's equation:
\[ x^2 + \left(\frac{x + 10}{2}\right)^2 = 100 \]
Expanding the equation:
\[ x^2 + \frac{(x + 10)^2}{4} = 100 \]
\[ x^2 + \frac{x^2 + 20x + 100}{4} = 100 \]
Multiply through by 4 to eliminate the fraction:
\[ 4x^2 + x^2 + 20x + 100 = 400 \]
Combine like terms:
\[ 5x^2 + 20x - 300 = 0 \]
Dividing the entire equation by 5:
\[ x^2 + 4x - 60 = 0 \]

### Step 2: Solve the quadratic equation for \( x \).
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-60)}}{2 \cdot 1} \]
\[ x = \frac{-4 \pm \sqrt{16 + 240}}{2} \]
\[ x = \frac{-4 \pm \sqrt{256}}{2} \]
\[ x = \frac{-4 \pm 16}{2} \]
Calculating the two possible values for \( x \):
1. \( x = \frac{12}{2} = 6 \)
2. \( x = \frac{-20}{2} = -10 \) (not in the first quadrant)

Thus, \( x = 6 \).

### Step 3: Find the corresponding \( y \) coordinate for point B.
Substituting \( x = 6 \) back into the line equation:
\[ y = \frac{6 + 10}{2} = \frac{16}{2} = 8 \]
So, the point \( B \) is \( (6, 8) \).

### Step 4: Find the equation of the line \( l_2 \) through point B, perpendicular to \( l_1 \).
The slope of line \( l_1 \) is \( \frac{1}{2} \) (from \( y = \frac{1}{2}x + 5 \)). The slope of \( l_2 \) will be the negative reciprocal:
\[ m_2 = -\frac{1}{\frac{1}{2}} = -2 \]
Using point-slope form for the line through \( B(6, 8) \):
\[ y - 8 = -2(x - 6) \]
Expanding this:
\[ y - 8 = -2x + 12 \implies y = -2x + 20 \]

### Step 5: Find the intersection points P and Q with the axes.
To find point \( P \) (where \( y = 0 \)):
\[ 0 = -2x + 20 \implies 2x = 20 \implies x = 10 \]
Thus, \( P(10, 0) \).

To find point \( Q \) (where \( x = 0 \)):
\[ y = -2(0) + 20 = 20 \]
Thus, \( Q(0, 20) \).

### Step 6: Calculate the area of triangle \( OPQ \).
The area \( A \) of triangle \( OPQ \) can be calculated using the formula for the area of a triangle:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base \( OP = 10 \) and the height \( OQ = 20 \):
\[ A = \frac{1}{2} \times 10 \times 20 = 100 \text{ sq. units} \]

### Final Answer:
The area of triangle \( OPQ \) is \( 100 \) square units.

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