Circles
Circle touching lines
Grade 11
Question:
<p><strong>Paragraph for Question nos. 624 and 625</strong><br>Let a line \(L_1\) passing through a point \(A(2, 0)\) and making an angle \(\theta\) with positive \(x\)-axis in anticlockwise direction, where \(\tan\theta = \frac{1}{2}\). Now, \(L_1\) is rotated about the point \(A\) in anticlockwise direction through an angle of \((\pi - 4\theta)\). If the line in new position is \(L_2\), then.<br><br>The radius of the largest circle which touches \(L_1\), \(L_2\) and the \(y\)-axis is:</p>
<p>\(4(\sqrt{3}-1)\)</p>
<p>\(4(\sqrt{3}+1)\)</p>
<p>\(4(\sqrt{5}-2)\)</p>
<p>\(4(\sqrt{5}+2)\)</p>
Step-by-Step Solution
Key Concept: Find equations of lines L₁ and L₂, then determine the circle touching both lines and the y-axis by using the property that the center lies on the angle bisector of the two lines and is equidistant from all three geometric objects.
<p><strong>Step 1: Find the equation of line L₁</strong></p><p>Line L₁ passes through A(2, 0) making angle θ with positive x-axis where tan θ = 1/2.</p><p>Equation of L₁: y - 0 = (1/2)(x - 2), which gives <strong>x - 2y - 2 = 0</strong></p><p><strong>Step 2: Find the angle of L₂</strong></p><p>L₁ is rotated about A through angle (π - 4θ) anticlockwise.</p><p>New angle = θ + (π - 4θ) = π - 3θ</p><p>tan(π - 3θ) = -tan(3θ)</p><p>Using tan(3θ) = (3tan θ - tan³θ)/(1 - 3tan²θ) with tan θ = 1/2:</p><p>tan(3θ) = (3(1/2) - (1/8))/(1 - 3/4) = (3/2 - 1/8)/(1/4) = (11/8)/(1/4) = 11/2</p><p>So tan(π - 3θ) = -11/2</p><p><strong>Step 3: Find equation of L₂</strong></p><p>Line L₂ passes through A(2, 0) with slope -11/2:</p><p>y - 0 = (-11/2)(x - 2), which gives <strong>11x + 2y - 22 = 0</strong></p><p><strong>Step 4: Set up circle tangent conditions</strong></p><p>Let the circle have center C(h, k) and radius r.</p><p>Since the circle touches the y-axis: <strong>h = r</strong></p><p>Distance from C to L₁: |h - 2k - 2|/√5 = r</p><p>Distance from C to L₂: |11h + 2k - 22|/√125 = r</p><p><strong>Step 5: Solve for h and k</strong></p><p>From condition 1: h = r</p><p>From condition 2: |r - 2k - 2|/√5 = r → |r - 2k - 2| = r√5</p><p>From condition 3: |11r + 2k - 22|/(5√5) = r → |11r + 2k - 22| = 5r√5</p><p>Taking r - 2k - 2 = r√5 (positive case, as circle is in the region where L₁ has positive distance):</p><p>2k = r - 2 - r√5 = r(1 - √5) - 2 ... (i)</p><p>Taking 11r + 2k - 22 = -5r√5 (circle lies on appropriate side):</p><p>2k = -5r√5 - 11r + 22 ... (ii)</p><p>From (i) and (ii):</p><p>r(1 - √5) - 2 = -5r√5 - 11r + 22</p><p>r - r√5 + 11r + 5r√5 = 24</p><p>r(12 + 4√5) = 24</p><p>r = 24/(12 + 4√5) = 6/(3 + √5)</p><p>Rationalizing: r = 6(3 - √5)/((3 + √5)(3 - √5)) = 6(3 - √5)/(9 - 5) = 6(3 - √5)/4 = 3(3 - √5)/2</p><p>r = (9 - 3√5)/2 = 4.5 - 1.5√5</p><p>Simplifying: r = 4(√5 - 2)</p><p><strong>∴ Answer: C</strong></p>
Correct Answer: C