<p>A circle passing through the origin and cutting equal chords of length \(\sqrt{2}\) from the straight line \(y = x\) and \(y = -x\)...</p>
<p>passes through (1, 0)</p>
<p>passes through (0, 1)</p>
<p>passes through (1, 1)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Since the circle passes through the origin and cuts equal chords from both y=x and y=-x, by symmetry the center must lie on a coordinate axis. Use the chord-distance formula: if a chord of length c is at distance d from center with radius r, then c² = 4(r² - d²).
<p><strong>Step 1:</strong> By symmetry of equal chords from y=x and y=-x, the center C(h,k) must satisfy h = ±k (or lie on coordinate axes). Since the circle passes through origin, r² = h² + k².</p><p><strong>Step 2:</strong> Distance from C(h,k) to line y=x is d₁ = |h-k|/√2, and to y=-x is d₂ = |h+k|/√2. For equal chord length √2: 2 = 4(r² - d₁²) = 4(r² - d₂²).</p><p><strong>Step 3:</strong> This gives d₁² = d₂², so |h-k|² = |h+k|². Expanding: h² - 2hk + k² = h² + 2hk + k², yielding hk = 0. Thus h=0 or k=0.</p><p><strong>Step 4:</strong> If h=0: r² = k², and 2 = 4(k² - k²/2) = 2k², so k=±1. If k=0: similarly h=±1. The circle is x² + y² ± 2y = 0 or x² + y² ± 2x = 0.</p><p>∴ Answer: A</p>
Correct Answer: A