Circles
Circumcircle of a Triangle
Grade 11
Question:
<p>Three sides of a triangle have the equations \(L_i \equiv y - m_i x = 0\); \(i = 1, 2\). Then \(L_1 L_2 + \lambda L_2 L_3 + \mu L_3 L_1 = 0\), where \(\lambda \neq 0, \mu \neq 0\), is the equation of the circumcircle of the triangle if</p>
<p>(a) \(1 + \lambda + \mu = m_1 m_2 + \lambda m_2 m_3 + \lambda m_3 m_1\)</p>
<p>(b) \(m_1(1 + \mu) + m_2(1 + \lambda) + m_3(\mu + \lambda) = 0\)</p>
<p>(c) \(\dfrac{1}{m_3} + \dfrac{1}{m_1} + \dfrac{1}{m_2} = 1 + \lambda + \mu\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: For a circumcircle passing through the vertices of a triangle formed by three lines, the combined equation L₁L₂ + λL₂L₃ + μL₃L₁ = 0 represents a circle only when the coefficients of x² and y² are equal and the xy term is zero. This requires specific relationships between λ and μ based on the slopes of the lines.
<p><strong>Step 1:</strong> Write the product of lines. The equation L₁L₂ + λL₂L₃ + μL₃L₁ = 0 represents a conic passing through the three vertices (intersection points of the lines).</p><p><strong>Step 2:</strong> Expand each product. With Lᵢ = y - mᵢx, we get products like (y - m₁x)(y - m₂x) = y² - (m₁ + m₂)xy + m₁m₂x².</p><p><strong>Step 3:</strong> Collect coefficients. After substitution and expansion, the coefficient of x² is: m₁m₂ + λm₂m₃ + μm₃m₁, and the coefficient of y² is: 1 + λ + μ.</p><p><strong>Step 4:</strong> Apply circle conditions. For a circle through three vertices:</p><ul><li>Coefficient of x² = Coefficient of y²: m₁m₂ + λm₂m₃ + μm₃m₁ = 1 + λ + μ</li><li>Coefficient of xy = 0: -(m₁ + m₂) - λ(m₂ + m₃) - μ(m₃ + m₁) = 0</li></ul><p><strong>Step 5:</strong> Solve the system. The xy-coefficient condition gives: m₁ + m₂ + λ(m₂ + m₃) + μ(m₃ + m₁) = 0. Combined with the equal coefficients condition, this determines the relationship between λ and μ in terms of m₁, m₂, m₃.</p><p>∴ Answer: B</p>
Correct Answer: B