JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

If the radius of the largest circle with centre $(2, 0)$ inscribed in the ellipse $x^2 + 4y^2 = 36$ is $r$ , then $12r^2$ is equal to:

Choose the correct answer:

Explanation

Solution:

Equation of the Circle: Let the circle be $(x - 2)^2 + y^2 = r^2$ .
Equation of the Ellipse: From $x^2 + 4y^2 = 36$ , we get $y^2 = \frac{36 - x^2}{4}$ .
Substitution: Substitute $y^2$ into the circle's equation:

$(x - 2)^2 + \frac{36 - x^2}{4} = r^2$

$4(x^2 - 4x + 4) + 36 - x^2 = 4r^2$

$3x^2 - 16x + (52 - 4r^2) = 0$
Condition for Tangency: For the largest inscribed circle, this quadratic in $x$ must have equal roots (Discriminant $D = 0$ ):

$D = (-16)^2 - 4(3)(52 - 4r^2) = 0$

$256 - 12(52 - 4r^2) = 0$

$256 - 624 + 48r^2 = 0$

$48r^2 = 368$

$12r^2 = 92$

Correct Option: (4) 92