Circles
Family of Circles
Grade 11
Question:
<p>In each of the following one or more options are correct. Choose the correct option(s).</p><p>(d) The circle of radius 1, touching the pair of lines \(12x^2 - 25xy + 12y^2 = 0\), \(x > 0\) has the equation</p>
<p>A. \(x^2 + y^2 + 10x + 10y + 49 = 0\)</p>
<p>B. \(x^2 + y^2 - 10x - 10y + 49 = 0\)</p>
<p>C. \(x^2 + y^2 + 10x - 10y + 49 = 0\)</p>
<p>D. none of these</p>
Step-by-Step Solution
Key Concept: A circle touching two lines must have its center equidistant from both lines, with this distance equal to the radius. First factorize the pair of lines, then use the distance formula from a point to a line.
<p><strong>Step 1:</strong> Factorize the pair of lines 12x² - 25xy + 12y² = 0</p><p>Using the quadratic formula in y: 12y² - 25xy + 12x² = 0</p><p>y = [25x ± √(625x² - 576x²)]/(24) = [25x ± 7x]/24</p><p>This gives: y = (4x/3) and y = (3x/4)</p><p>So the lines are: 4x - 3y = 0 and 3x - 4y = 0</p><p><strong>Step 2:</strong> The angle bisectors of these lines are found by:</p><p>(4x - 3y)/5 = ±(3x - 4y)/5</p><p>This gives: x = y and x + 7y = 0</p><p><strong>Step 3:</strong> For x > 0, the center lies on x = y (the bisector keeping x > 0)</p><p>Let center be (h, h) where h > 0</p><p><strong>Step 4:</strong> Distance from (h, h) to line 4x - 3y = 0 equals radius 1:</p><p>|4h - 3h|/5 = 1</p><p>|h|/5 = 1 ⟹ h = 5</p><p><strong>Step 5:</strong> The center is (5, 5) and radius is 1</p><p>∴ Answer: (x - 5)² + (y - 5)² = 1</p>
Correct Answer: B