<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>The point of intersection of the lines FA and BC is:</p>
<p>(a) N</p>
<p>(b) outside S₁ but inside S₂</p>
<p>(c) outside S₂ but inside S₁</p>
<p>(d) inside S₁ and S₂ both</p>
Step-by-Step Solution
Key Concept: The key is recognizing that lines FA and BC, being lines through opposite vertices of the circumcircles, intersect at the second common point N of the two circles.
<p><strong>Step 1:</strong> Let M = (m, 0) where 0 < m < 4. Square AMCD has vertices A(0,0), M(m,0), C(m,m), D(0,m). Square BMFE has vertices B(4,0), M(m,0), F(m, 4-m), E(4, 4-m).</p><p><strong>Step 2:</strong> The circumcircle S₁ of square AMCD has centre P at (m/2, m/2) and radius \(\frac{m\sqrt{2}}{2}\). The circumcircle S₂ of square BMFE has centre Q at ((4+m)/2, (4-m)/2) and radius \(\frac{(4-m)\sqrt{2}}{2}\).</p><p><strong>Step 3:</strong> Line FA passes through F(m, 4-m) and A(0, 0). Line BC passes through B(4, 0) and C(m, m).</p><p><strong>Step 4:</strong> Using properties of radical axes and the fact that both S₁ and S₂ pass through M and N, the lines FA and BC intersect at the second intersection point N of the two circles.</p><p>∴ Answer is (a).</p>
Correct Answer: A