Let S₁ : x² + y² = 9 and S₂ : (x - 2)² + y² = 1. Then the locus of center of a variable circle S which touches S₁ internally and S₂ externally always passes through the points :

Solution

Maths sir · 4 steps

Write both touching conditions as distances and add them: the radius cancels and an ellipse appears.

1. Centres and radii. S₁ has centre O(0, 0) and radius 3, and S₂ has centre C(2, 0) and radius 1. Let S have centre P and radius r.
2. Touching conditions. Touching S₁ internally gives OP = 3 - r. Touching S₂ externally gives PC = 1 + r.

   OP + PC = (3 - r) + (1 + r) = 4

3. An ellipse. A constant sum of distances to (0, 0) and (2, 0) means P lies on an ellipse with these two points as foci, so 2a = 4 and the centre is (1, 0). With a = 2 and c = 1, b² = a² - c² = 3, so the locus is (x-1)²/4 + y²/3 = 1.

   Option	Point	(x-1)²/4 + y²/3	On the locus?
   (a)	(1/2, ±√5/2)	1/16 + 5/12 = 23/48	no
   (b)	(1, ± 2)	0 + 4/3 = 4/3	no
   (c)	(2, ±3/2)	1/4 + 3/4 = 1	yes
   (d)	(0, ±√3)	1/4 + 1 = 5/4	no

4. So. Only (2, ±3/2) satisfies the equation. As a check, at (2, 3/2) we get OP = 5/2 and PC = 3/2, which add to 4 with r = 1/2. The answer is (c).

The idea to keep. When a circle touches one circle internally and another externally, add the two centre distances. The unknown radius cancels and the locus is an ellipse with the two given centres as its foci.

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