JEE Main 2026 · 2 Apr · Shift 2 · Mathematics Q23

← JEE Main 2026 · 2 Apr · Shift 2

MathematicsInteger+4 / -1Question 23 of 75

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 ________.

Your answer

Answer and worked solution

Official answer: 18

Tangency condition. The circle on AB has centre $M=\frac{A+B}2$ and radius $\frac{AB}2$ . Internal tangency to $x^2+y^2=36$ means $OM=6-\frac{AB}2$ , i.e. $|A+B|+|B-A|=12$ .

Locus. $|A+B|$ is the distance from B to $(-3,0)$ and $|B-A|$ the distance to $(3,0)$ . So B lies on the ellipse with foci $(\pm3,0)$ and $2a=12$ : $e=\frac36=\frac12$ .

Value. $72e^2=18$ .

Answer: 18

Topics:Ellipse Tangents and normals Locus

← Previous question Next question →

JEE Main 2026 · 2 Apr · Shift 2, Mathematics question 23