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Locus with Internal Tangency is an Ellipse: 72e^2 = 18 | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
Let AA be the point (3,0)(3,0) and circles with variable diameter ABAB touch the circle x2+y2=36x^2+y^2=36 internally. Let the curve CC be the locus of the point BB. If the eccentricity of CC is ee, then 72e272e^2 is equal to
Solution
Answer: 18 (± 0.01)
Step 1: Let B(h,k)B(h,k). The circle on diameter ABAB (with A(3,0)A(3,0)) is
(x−h)(x−3)+(y−k)(y−0)=0 ⇒ x2+y2−(h+3)x−ky+3h=0,(x-h)(x-3)+(y-k)(y-0)=0\ \Rightarrow\ x^2+y^2-(h+3)x-ky+3h=0,
 with centre (h+32,k2)\left(\dfrac{h+3}{2},\dfrac{k}{2}\right) and some radius r1r_1. Step 2: Internal tangency with the fixed circle (centre OO, radius 66): distance between centres =∣6−r1∣=|6-r_1|. Working this out (the variable circle passes through AA and BB, radius =12AB=\tfrac12 AB) gives the condition
(h+3)2+k2+(h−3)2+k2=12.\sqrt{(h+3)^2+k^2}+\sqrt{(h-3)^2+k^2}=12.
 Step 3: This is the definition of an ellipse: the sum of distances from BB to the two fixed points (−3,0)(-3,0) and (3,0)(3,0) is constant =12=2a=12=2a, so a=6a=6, and the foci are (±3,0)(\pm3,0). Step 4: For the ellipse, 2ae=2ae= distance between foci =6=6:
12e=6 ⇒ e=12.12e=6\ \Rightarrow\ e=\frac12.
 Step 5:
72e2=72⋅14=18.72e^2=72\cdot\frac14=18.
 Correct answer: 18
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