Q.

Let C:x2+y2=4 and C':x2+y2–4λx+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 (8a + 12, 16b – 20) lies on the curve :          [2024]

1 x2–4y2=7  
2 6x2+y2=42  
3 5x2–y=–11  
4 x2+2y2–5x+6y=3  

Ans.

(2)

We have, C : x2+y2=4                         ... (i)

and C' : x2+y2–4λx+9=0

⇒ C' : (x–2λ)2+y2=4λ2–9           ... (ii)

Radius of C, r1=2 and radius of C', r2=4λ2–9

When two circles intersect at two points,

then |r1–r2|<CC'<|r1+r2|

⇒|2–4λ2–9|<2λ<2+4λ2–9              ...(iii)

By (iii), we have 4+4λ2–9–44λ2–9<4λ2

⇒ –5–44λ2–9<0

⇒ 5+44λ2–9>0 ⇒ 4λ2–9>–54

On Squaring both sides, we get

4λ2–9>2516 ⇒ 4λ2>16916 ⇒ λ2>16964

⇒ λ>138  or  λ<–138

∴ λ∈(–∞,–138)∪(138,∞)

Also, 4λ2<(2+4λ2–9)2           [By (iii)]

⇒ 4λ2<4+44λ2–9+4λ2–9

⇒ 0<–5+44λ2–9 ⇒ 5<44λ2–9

On squaring, we get

2516<4λ2–9 ⇒ 16964<λ2

∴ λ∈(–∞,–138)∪(138,∞)

Thus, Circles C and C' intersect at two distinct points for

λ∈R–[–138,138]

∴ a=–138, b=138

∴  (8a + 12, 16b –20) = (–1, 6) which satisfies only 6x2+y2= 42.