Let C: x2+y2=4 and C': x2+y2-4 x+9=0 be two circles. If the set of all values of 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 :
- x2+2y2-5x+6y=3
- 5x2-y=-11
- x2-4y2=7
- 6x2+y2=42
Answer
(4) 6x2+y2=42
Solution
C: centre (0, 0), r1=2; C': centre (2 , 0), r2= 4 2-9, distance of centres d=2| |
∵ two distinct points of intersection:
|r1-r2|<d<r1+r2
d<r1+r2: 4 2<4+(4 2-9)+4 4 2-9
4 2-9> 54 2> 16964 | |> 138
d>|r1-r2|: 4 2>4+(4 2-9)-4 4 2-9 4 4 2-9>-5, always true
∴ - [- 138, 138 ] a=- 138, b= 138
Point =(-13+12, 26-20)=(-1, 6)
6(-1)2+62=42 lies on 6x2+y2=42
Answer: (4)