<p><strong>Paragraph for Question nos. 652 and 653</strong><br>Let \(A\) be a variable point on locus of feet of perpendicular drawn from focus upon any tangent to the curve \(|z-2|+|z+2|=6\) and \(B\) be a variable point on \((1-i)z + (1+i)\bar{z} = 10\sqrt{2}\), then<br><br>If a variable circle touches both the loci on which \(A\) and \(B\) externally lie then latus rectum of locus of centre of variable circle is:</p>
Step-by-Step Solution
Key Concept: The locus of feet of perpendiculars from a focus to tangents of an ellipse is the auxiliary circle. For the ellipse |z-2|+|z+2|=6, this gives a circle of radius 3 centered at origin. The line containing B is tangent to a circle, and the locus of centers of circles touching both loci externally forms a hyperbola whose latus rectum we need to find.
<p><strong>Step 1:</strong> Identify the ellipse from |z-2|+|z+2|=6. Foci are at (±2,0), and 2a=6 so a=3, c=2, b²=9-4=5.</p><p><strong>Step 2:</strong> The locus of feet of perpendiculars from focus to tangents of an ellipse is the auxiliary circle: |z|=3 (radius a=3, centered at origin).</p><p><strong>Step 3:</strong> Convert the line (1-i)z+(1+i)z̄=10√2. Substituting z=x+iy: this simplifies to x+y=5√2. This line is tangent to circle |z|=r₀ where r₀=5√2/√2=5.</p><p><strong>Step 4:</strong> Let the variable circle have center C and radius r. If it touches the auxiliary circle |z|=3 externally: |C|=3+r. If it touches the line externally: distance from C to line = r, giving |x_C+y_C-5√2|/√2=r.</p><p><strong>Step 5:</strong> From touching conditions: |C|=3+r and r=|x_C+y_C-5√2|/√2. This gives |C|-3=(distance from C to line)=r, leading to |C|-(distance from C to line)=3, which is a hyperbola with difference of distances=2a'=3, so a'=3/2.</p><p><strong>Step 6:</strong> The directrix (line) is at distance 5 from origin. For hyperbola: c'=5, a'=3/2, so b'²=c'²-a'²=25-9/4=91/4.</p><p><strong>Step 7:</strong> Latus rectum of hyperbola = 2b'²/a'=(2×91/4)/(3/2)=(91/2)×(2/3)=91/3.</p><p>∴ Answer: C</p>
Correct Answer: C