Aspire Faculty ID #18029 · Topic: JEE Main 2026 (23 January Evening Shift) · Just now
JEE Main 2026 (23 January Evening Shift)

An equilateral triangle $OAB$ is inscribed in the parabola $y^2 = 4x$ with vertex at origin. Then the minimum distance of the circle having $AB$ as diameter from origin is:

Solution


Let $A(at^2, 2at)$

$t = 2\sqrt{3}$

Required circle:

$(x - 12)^2 + y^2 = (4\sqrt{3})^2$

Minimum distance

$= |CP - R| = 4(3 - \sqrt{3})$


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