Let C: x²+y²=4 and C′: x²+y²-4 λ x+9=0 be two circles. If the set of all values of λ so that the circles C and C intersect at two distinct points, is R-[a, b], then the point (8 a+12,16 b-20) lies on the curve :

Solution

Maths sir · 4 steps

Subtract the two circles to get the common chord, then make that line cut C at two points.

1. Common chord. Any common point satisfies both equations. Subtract C from C': -4λ x + 9 = -4. So every common point lies on one vertical line.

   x = 13/4λ

2. Two distinct points. A vertical line x = k cuts x² + y² = 4 at two distinct points only when |k| < 2. So |13/4λ| < 2, which gives |λ| > 13/8. This also makes 4λ² - 9 > 0, so C' is a real circle.
3. Find the point. The allowed set is ℝ - [-13/8, 13/8], so a = -13/8 and b = 13/8. Then 8a + 12 = -13 + 12 = -1 and 16b - 20 = 26 - 20 = 6, so the point is (-1, 6).

   Option	Left side at (-1, 6)	Right side	Lies on it?
   (a)	1 + 72 + 5 + 36 = 114	3	no
   (b)	5 - 6 = -1	-11	no
   (c)	1 - 144 = -143	7	no
   (d)	6 + 36 = 42	42	yes

4. So. The point (-1, 6) lies on 6x² + y² = 42, which is option (d).

The idea to keep. Two circles meet where their common chord meets either circle. Subtract the equations to get the chord, then demand that the distance from the centre to the chord is less than the radius.

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