If the radius of the largest circle with centre (2,0) inscribed in the ellipse x²+4y²=36 is r, then 12r² is equal to :

Solution

Maths sir · 4 steps

The largest circle just touches the ellipse, so r is the shortest distance from (2, 0) to the ellipse.

1. Why shortest distance. A circle centred at (2, 0) stays inside the ellipse only while its radius is at most the distance to the nearest point of the ellipse. The largest such circle touches the ellipse at that nearest point, so r is this minimum distance.
2. Distance squared. The ellipse is x²/36 + y²/9 = 1, so y² = 9 - x²/4 with -6 ≤ x ≤ 6. For a point (x, y) on it, the squared distance from (2, 0) is (x-2)² + y².

   d² = (x-2)² + 9 - x²/4 = 3/4x² - 4x + 13

3. Minimise. This is an upward parabola in x, so its least value is at x = 4/(2 × 3/4) = 8/3. That lies inside [-6, 6], so the point is on the ellipse. There, d² = 3/4 × 64/9 - 32/3 + 13 = 16/3 - 32/3 + 13 = 23/3.
4. So. r² = 23/3, so 12r² = 12 × 23/3 = 92, which is option (b).

The idea to keep. Largest circle inside a curve with a fixed centre means minimum distance from that centre to the curve. Write the distance squared in one variable, minimise it, and check that the point actually lies on the curve.

Answer from the NTA final answer key. Solution by PrepWiser.