Q96 / 180 · KCET 2007 · Mathematics

$x^{2}+y^{2}-6 x-6 y+4=0, \quad x^{2}+y^{2}-2 x$ $-4 y+3=0, x^{2}+y^{2}+2 k x+2 y+1=0$. If the radical centre of the above three circles exists, then which of the following cannot be the value of $k$ ?

  • A
    2
  • B
    1
  • Correct answer
    C
    5
  • D
    4

Solution

Given equation of circles are $S_{1} \equiv x^{2}+y^{2}-6 x-6 y+4=0$ $\quad S_{2} \equiv x^{2}+y^{2}-2 x-4 y+3=0$ and $S_{3} \equiv x^{2}+y^{2}+2 k x+2 y+1=0$ Now, radical axis of circle $S_{1}$ and $S_{2}$ is $$ S_{1}-S_{2}=0 $$ $\Rightarrow x^{2}+y^{2}-6 x-6 y+4-x^{2}-y^{2}+2 x$ $\Rightarrow \quad-4 x-2 y+1=0$ $\Rightarrow \quad 4 x+2 y-1=0$ Radical axis of circle $S_{2}$ and $S_{3}$ is $-3=0$ $\quad S_{2}-S_{3}=0$ $\Rightarrow x^{2}+y^{2}-2 x-4 y+3-x^{2}-y^{2}-2 k x$ $\Rightarrow \quad-(2+2 k) x-6 y+2=0$ $\Rightarrow \quad(2+2 k) x+6 y-2=0$ $$ \begin{array}{lr} \Rightarrow & 24-4-4 k \neq 0 \\ \Rightarrow & 20-4 k \neq 0 \\ \Rightarrow & k \neq 5 \end{array} $$

Hence, the correct answer is option: 'C'.

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