If the radius of the largest circle with centre (2, 0) inscribed in the ellipse x 2 + 4 y 2 = 36 is r , then 12 r 2 is equal to [2023]

Question

If the radius of the largest circle with centre (2, 0) inscribed in the ellipse $x^{2} + 4 y^{2} = 36$ is $r$ , then $12 r^{2}$ is equal to [2023]

Smart Achievers · Accepted Answer

(4)

Centre of the circle is (2, 0).

Ellipse : $x^{2} + 4 y^{2} = 36$

$\frac{x^{2}}{36} + \frac{y^{2}}{9} = 1$ $\dots (i)$

The equation of the circle is, ${(x - 2)}^{2} + y^{2} = r^{2}$ $\dots (i i)$

Solving (i) and (ii), we get

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

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

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

Discriminant = 0

$∴$ ${(16)}^{2} - 4 \times 3 \times (52 - 4 r^{2}) = 0$

$\Rightarrow 52 - 4 r^{2} = \frac{256}{12}$ $\Rightarrow 4 r^{2} = 52 - \frac{256}{12} = \frac{368}{12}$

$\Rightarrow 12 r^{2} = \frac{368}{4} = 92$

Solution

Q.
If the radius of the largest circle with centre (2, 0) inscribed in the ellipse $x^{2} + 4 y^{2} = 36$ is $r$ , then $12 r^{2}$ is equal to [2023]

1 72

2 69

3 115

4 92

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