<p>A circle passing through the point \((3, \sqrt{2})\) touches the pair of lines \(x^2 - y^2 - 2x + 1 = 0\). The centre of the circle is</p>
Step-by-Step Solution
Key Concept: The pair of lines factors as (x-y-1)(x+y-1)=0, forming two lines. The center of any circle touching both lines must lie on the angle bisector of these lines, equidistant from both.
<p><strong>Step 1:</strong> Factor the pair of lines: x² - y² - 2x + 1 = 0 becomes (x-1)² - y² = 0, which gives (x-y-1)(x+y-1) = 0</p><p>So the two lines are: L₁: x - y - 1 = 0 and L₂: x + y - 1 = 0</p><p><strong>Step 2:</strong> For a circle to touch both lines, its center must lie on an angle bisector. The angle bisectors are found by: (x-y-1)/√2 = ±(x+y-1)/√2</p><p>This gives: x - y - 1 = ±(x + y - 1)</p><p>Taking positive sign: -2y = 0 → y = 0</p><p>Taking negative sign: -2x + 2 = 0 → x = 1</p><p><strong>Step 3:</strong> The center lies on y = 0 or x = 1. Let center be (h, k). Since circle passes through (3, √2), the radius r satisfies: r² = (h-3)² + (k-√2)²</p><p><strong>Step 4:</strong> Distance from (h,k) to line L₁: |h-k-1|/√2 = r, and to L₂: |h+k-1|/√2 = r</p><p>If center is at (h, 0): |h-1|/√2 = r and |h-1|/√2 = r ✓ (consistent)</p><p>Then: (h-1)²/2 = (h-3)² + 2</p><p>(h-1)² = 2(h-3)² + 4</p><p>h² - 2h + 1 = 2h² - 12h + 18 + 4</p><p>h² - 10h + 21 = 0 → (h-3)(h-7) = 0</p><p>∴ Center is (3, 0) or (7, 0)</p><p><strong>Answer: A</strong></p>
Correct Answer: A