Question

If the angle between the pair of straight lines formed by joining the points of intersection of \(x^2 + y^2 = 4\) and \(y = 3x + c\) to the origin is a right angle, then \(c^2\) is:

• A. \(20\)
• B. \(13\)
• C. \( \frac{1}{5} \)
• D. \(5\)

Hint:

Use product of slopes = -1 for perpendicular lines.

Answer 1

The Correct Option is A

Concept: Condition for perpendicular lines: \[ m_1 m_2 = -1 \]

Step 1:Substitute \(y = 3x + c\) in circle: \[ x^2 + (3x + c)^2 = 4 \] \[ x^2 + 9x^2 + 6cx + c^2 = 4 \Rightarrow 10x^2 + 6cx + (c^2 - 4) = 0 \]

Step 2:Slopes of lines from origin: \[ m = \frac{y}{x} = 3 + \frac{c}{x} \] Using quadratic roots: \[ x_1 x_2 = \frac{c^2 - 4}{10} \]

Step 3:Condition: \[ m_1 m_2 = -1 \Rightarrow \frac{(3x_1 + c)(3x_2 + c)}{x_1 x_2} = -1 \] Solving gives: \[ c^2 = 20 \]