Circles
Family of circles through intersection of lines
Grade 11
Question:
<p>If \(L_1, L_2, L_3\) are three lines, then \(L_1 L_2 + \lambda L_2 L_3 + \mu L_3 L_1 = 0\) represents a circle if which of the following conditions hold?</p>
<p>(a) \(1 + \lambda + \mu = m_1 m_2 + \lambda m_2 m_3 + \mu m_1 m_3\)</p>
<p>(b) \(m_1(1+\mu) + m_2(1+\lambda) + m_3(\mu+\lambda) = 0\)</p>
<p>(c) coefficient of \(x^2\) = coefficient of \(y^2\)</p>
<p>(d) coefficient of \(xy = 0\)</p>
Step-by-Step Solution
Key Concept: For a pair of lines to represent a circle when combined in the form L₁L₂ + λL₂L₃ + μL₃L₁ = 0, the three lines must be concurrent (meet at a point) and the coefficients must satisfy λ = μ = 1, making it a pair of lines through the point of concurrency. The 'circle' is actually a degenerate case where the pair of angle bisectors of the lines are perpendicular.
<p><strong>Step 1:</strong> Recognize that L₁L₂ + λL₂L₃ + μL₃L₁ = 0 is a product of three line equations, which represents a pair of lines (degree 2 equation).</p><p><strong>Step 2:</strong> For this expression to represent a circle, we need the pair of lines to be imaginary conjugates, which occurs when the three lines L₁, L₂, L₃ are concurrent (meet at a single point).</p><p><strong>Step 3:</strong> When three concurrent lines are given and we form L₁L₂ + λL₂L₃ + μL₃L₁ = 0, this represents a pair of lines through the point of concurrency.</p><p><strong>Step 4:</strong> For these two lines to form a circle (degenerate), they must be perpendicular bisectors of angles formed by the original lines. This requires λ = μ = 1, giving L₁L₂ + L₂L₃ + L₃L₁ = 0.</p><p><strong>Step 5:</strong> The additional condition is that the angle bisectors of any two pairs from {L₁, L₂, L₃} must be perpendicular.</p><p>∴ Answer: The conditions are (a) the three lines are concurrent AND (b) λ = μ = 1</p>
Correct Answer: a, b