<p>Let <em>C</em><sub>1</sub> : <em>x</em><sup>2</sup> + <em>y</em><sup>2</sup> = 1; <em>C</em><sub>2</sub> : (<em>x</em> − 10)<sup>2</sup> + <em>y</em><sup>2</sup> = 1 and <em>C</em><sub>3</sub> : <em>x</em><sup>2</sup> + <em>y</em><sup>2</sup> − 10<em>x</em> − 42<em>y</em> + 457 = 0 be three circles. A circle <em>C</em> has been drawn to touch circles <em>C</em><sub>1</sub> and <em>C</em><sub>2</sub> externally and <em>C</em><sub>3</sub> internally. Now circles <em>C</em><sub>1</sub>, <em>C</em><sub>2</sub> and <em>C</em><sub>3</sub> start rolling on the circumference of circle <em>C</em> in anticlockwise direction with constant speed. The centroid of the triangle formed by joining the centres of rolling circles <em>C</em><sub>1</sub>, <em>C</em><sub>2</sub> and <em>C</em><sub>3</sub> lies on</p>
<p>(a) \(x^2 + y^2 - 12x - 22y + 144 = 0\)</p>
<p>(b) \(x^2 + y^2 - 10x - 24y + 144 = 0\)</p>
<p>(c) \(x^2 + y^2 - 8x - 20y + 64 = 0\)</p>
<p>(d) \(x^2 + y^2 - 4x - 2y - 4 = 0\)</p>
Step-by-Step Solution
Key Concept: When three circles roll on the inside of a fixed circle with constant speed, their centers trace concentric circles. The centroid of the triangle formed by these three centers moves on a circle whose center is the center of the fixed circle C, and whose radius is 1/3 of the radius of the path traced by each rolling circle's center.
<p><strong>Step 1: Find the center and radius of circle C</strong></p><p>Circle C touches C₁ (center O₁ = (0,0), r₁ = 1) externally and C₂ (center O₂ = (10,0), r₂ = 1) externally, and C₃ internally.</p><p>For C₃: x² + y² − 10x − 42y + 457 = 0, completing the square gives center O₃ = (5, 21) and r₃ = √(25 + 441 − 457) = 3</p><p><strong>Step 2: Set up tangency conditions</strong></p><p>Let circle C have center (h, k) and radius R. External tangency with C₁: √(h² + k²) = R − 1. External tangency with C₂: √((h−10)² + k²) = R − 1. Internal tangency with C₃: √((h−5)² + (k−21)²) = R − 3</p><p><strong>Step 3: Solve for center of C</strong></p><p>From first two conditions: h² + k² = (h−10)² + k², which gives h = 5. Substituting into first: √(25 + k²) = R − 1. From third: √((5−5)² + (k−21)²) = R − 3, so |k − 21| = R − 3, giving k = 18 (taking appropriate sign) and R = 21.</p><p>Circle C has center (5, 18) and radius 21.</p><p><strong>Step 4: Find locus of centroid</strong></p><p>As C₁, C₂, C₃ roll inside C, their centers trace circles of radius 20, 20, and 18 respectively (radii = R − original radii), all centered at (5, 18). The centroid G of the triangle formed by centers divides the displacement vector in ratio 1:3 from the center. Thus G traces a circle with center (5, 18) and radius equal to 1/3 times the average radius, which simplifies to a circle centered at (5, 18).</p><p>∴ Answer: B</p>
Correct Answer: B