Q81 / 180 · COMEDK 2020 · Mathematics
$x^{2}+y^{2}-6 x-6 y+4=0$,
$x^{2}+y^{2}-2 x-4 y+3=0$,
$x^{2}+y^{2}+2 k x+2 y+1=0$. If the Radical centre of the above three circles exists, then which of the following cannot be the value of $k$ ?
- A1
- B2
- C4
- Correct answerD5
Solution
Let
$$
\begin{aligned}
&S_{1}: x^{2}+y^{2}-6 x-6 y+4=0 \\
&S_{2}: x^{2}+y^{2}-2 x-4 y+3=0 \\
&S_{3}: x^{2}+y^{2}+2 k x+2 y+1=0
\end{aligned}
$$
Now, radical cnetre of $S_{1}$ and $S_{\text {g }}$ is given by
$$
\begin{aligned}
& & \mathrm{S}_{1}-\mathrm{S}_{2} &=0 \\
\Rightarrow & &-4 x-2 y+1 &=0 \\
\Rightarrow & & 4 x+2 y-1 &=0
\end{aligned}
$$
Radical centre of $S_{2}$ and $S_{3}$ is given by
$$
\begin{array}{r}
\mathrm{S}_{2}-\mathrm{S}_{3}=0 \\
-2(1+k) x-6 y+2=0 \\
(1+k) x+3 y-1=0
\end{array}
$$
Now, Eq. (ii) becomes, if
$$
k=1: 2 x+3 y-1=0
$$
$$
\begin{aligned}
&k=2: 3 x+3 y-1=0 \\
&k=4: 5 x+3 y-1=0 \\
&k=5: 6 x+3 y-1=0
\end{aligned}
$$
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, $k \neq 5$
Hence, the correct answer is option: 'D'.