x²+y²-6 x-6 y+4=0 , x²+y²-2 x-4 y+3=0 , x²+y²+2 k x+2 y+1=0 . If the Radical centre of the above three circles exists, then which of the following cannot be the value of k ?

Step-by-step solution

Let aligned &S₁: x²+y²-6 x-6 y+4=0 &S₂: x²+y²-2 x-4 y+3=0 &S₃: x²+y²+2 k x+2 y+1=0 aligned Now, radical cnetre of S₁ and S_ g is given by aligned & & S ₁- S ₂ &=0 & &-4 x-2 y+1 &=0 & & 4 x+2 y-1 &=0 aligned Radical centre of S₂ and S₃ is given by array r S ₂- S ₃=0 -2(1+k) x-6 y+2=0 (1+k) x+3 y-1=0 array Now, Eq. (ii) becomes, if k=1: 2 x+3 y-1=0 aligned &k=2: 3 x+3 y-1=0 &k=4: 5 x+3 y-1=0 &k=5: 6 x+3 y-1=0 aligned Now, only 6 x+3 y-1=0 is parallel to 4 x+2 y-1=0 So, for this radical centre does not exists. So, k 5