<p>The equation of a circle which touches the line \(2x - y = 1\) at (1, 1) and also touches the line \(2x + y = 4\) is</p>
<p>\(x^2 + y^2 + 2y + 1 = 0\)</p>
<p>\(x^2 + y^2 + 3y + 1 = 0\)</p>
<p>\(x^2 + y^2 - 3y + 1 = 0\)</p>
<p>\(4(x^2 + y^2) + 10x - 7y - 9 = 0\)</p>
Step-by-Step Solution
Key Concept: The center of a circle must lie on the perpendicular to any tangent line at the point of tangency. Since the circle touches both lines, its center lies on perpendiculars to both lines, and the distances from center to both lines equal the radius.
<p><strong>Step 1:</strong> Find the perpendicular to 2x - y = 1 at point (1,1).</p><p>The line 2x - y = 1 has slope 2, so the perpendicular has slope -1/2.</p><p>Perpendicular through (1,1): y - 1 = -½(x - 1) ⟹ x + 2y = 3</p><p><strong>Step 2:</strong> The center lies on x + 2y = 3. Let center be (h, k) where h + 2k = 3, so h = 3 - 2k.</p><p><strong>Step 3:</strong> Distance from center to tangent line 2x - y = 1 equals radius:</p><p>r = |2h - k - 1|/√5 = |2(3 - 2k) - k - 1|/√5 = |5 - 5k|/√5 = |5(1 - k)|/√5</p><p><strong>Step 4:</strong> Distance from center to line 2x + y = 4 also equals radius:</p><p>r = |2h + k - 4|/√5 = |2(3 - 2k) + k - 4|/√5 = |2 - 3k|/√5</p><p><strong>Step 5:</strong> Equate the two expressions for r:</p><p>|5(1 - k)| = |2 - 3k|</p><p>Case 1: 5(1 - k) = 2 - 3k ⟹ 5 - 5k = 2 - 3k ⟹ k = 3/2</p><p>Then h = 3 - 2(3/2) = 0, so center is (0, 3/2) and r = |5(1 - 3/2)|/√5 = 5/(2√5) = √5/2</p><p><strong>Step 6:</strong> Equation of circle: x² + (y - 3/2)² = 5/4</p><p>Expanding: x² + y² - 3y + 9/4 = 5/4 ⟹ x² + y² - 3y + 1 = 0</p><p>∴ Answer: C</p>
Correct Answer: C