JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Ellipse

If the radius of the largest circle with centre

(2, 0)

inscribed in the ellipse

x^{2} + 4 y^{2} = 36

r

, then

12 r^{2}

is equal to

Answer: (C)

Solution

The circle has equation

(x - 2)^{2} + y^{2} = r^{2}

. For the circle to be inscribed (tangent to the ellipse), substitute

y^{2} = \frac{36 - x ^{2}}{4}

from the ellipse equation:

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

Expanding and simplifying:

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

4 x^{2} - 16 x + 16 + 36 - x^{2} = 4 r^{2}

3 x^{2} - 16 x + 52 - 4 r^{2} = 0

For tangency, the discriminant must be zero:

D = 256 - 12 (52 - 4 r^{2}) = 0

256 = 624 - 48 r^{2}

48 r^{2} = 368

4 r^{2} = \frac{92}{3}

12 r^{2} = 92

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Single correct option