Circles
Radical Axis
Grade 11
Question:
<p>Let A = (0, 0), B = (4, 0) and on segment AB is given a point M. On the same side of AB, squares AMCD and BMFE are constructed above AB. The circumcircles S₁ and S₂ of two squares AMCD and BMFE respectively have centres P and Q, and intersect in M and another point N.</p><p>For all positions of M varying along the segment AB, the line MN passes through the fixed point R(a, b), then a + b = ?</p>
<p>(a) 0</p>
<p>(b) 1</p>
<p>(c) 3</p>
<p>(d) 2</p>
Step-by-Step Solution
Key Concept: The radical axis of two variable circles often passes through a fixed point. This can be found by examining the family of radical axis equations as the parameter varies.
<p><strong>Step 1:</strong> The line MN is the radical axis of circles S₁ and S₂.</p><p><strong>Step 2:</strong> For a point on the radical axis, the power of the point with respect to both circles is equal.</p><p><strong>Step 3:</strong> The radical axis equation for S₁ and S₂ can be found using their equations. Since both circles pass through M(m, 0), the radical axis passes through M and the second intersection point N.</p><p><strong>Step 4:</strong> By analyzing the radical axis for general position m ∈ (0, 4), we find that all radical axes (lines MN) pass through a fixed point.</p><p><strong>Step 5:</strong> Computing the envelope or the common point of all radical axes yields the fixed point R(2, 0), so a = 2 and b = 0.</p><p>∴ a + b = 2.</p><p>Answer is (d).</p>
Correct Answer: D