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JEE Main20262 April 2026Evening ShiftMathematicsEllipseActual

Let A be the point (3, 0) and circles with variable diameter AB touch the circle x^2 + y^2 = 36 internally. Let the curve C be the locus of the point B. If the eccentricity of C is e , then 72e^2 is equal to _______.

Correct answer

0

Step-by-step solution

Let the coordinates of point B be (h, k) . The center of the circle with diameter AB is C₁ ( h+3 2 , k 2 ) and its radius is r₁ = 1 2 (h-3)^2 + k^2 . The given circle is x^2 + y^2 = 36 , which has center C₂(0, 0) and radius r₂ = 6 . Since the circles touch internally, the distance between their centers is equal to the difference of their radii: C₁C₂ = r₂ - r₁ ( h+3 2 )^2 + ( k 2 )^2 = 6 - 1 2 (h-3)^2 + k^2 Multiplying the entire equation by 2, we get: (h+3)^2 + k^2 + (h-3)^2 + k^2 = 12 This equation represents the

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