Question

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

• A. 1
• B. 2
• C. 4
• D. 5

Answer 1

Answer: (D)

Given, equation of circles are
$S_1 = x^2 + y^2 - 6 x - 6 y + 4 = 0$
$S_2 = x^2 + y^2- 2 x — 4 y + 3 = 0$
and $S_3 = x^2 + y^2 + 2kx + 2y + 1 = 0$
Now, radical axis of circle $S_1$ and $S_2$ is
$S_1 — S_2 = 0$
$ \Rightarrow \, x^2 + y^2 - 6x - 6y + 4 - x^2- y^2+ 2 x$
$ \Rightarrow \, - 4 x - 2 y + 1 = 0$
$ \Rightarrow \, 4 x + 2 y + l = 0\,\,\,\,\,\dots(i)$
Radical axis of circle $S_2$ and $S_3$ is
$S_2 — S_3 = 0$
$ \Rightarrow \, x^2 + y^2- 2 x - 4 y + 3 x^2 - y^2 - 2kx$
$ \Rightarrow \, - (2 + 2k) x - 6y + 2 = 0$
$ \Rightarrow \, (2 + 2k) x + 6 y - 2 = 0\,\,\,\,\,\dots(ii)$
For existence of radical centre
$\begin{vmatrix}4&2\\ 2+2k&6\end{vmatrix} \ne 0$
$ \Rightarrow \, 24 - 2 (2 + 2k) \ne 0$
$ \Rightarrow \, 2 4 - 4 - 4 k \ne 0 $
$ \Rightarrow \, 20 - 4 k \ne 0$
$ \Rightarrow \, k \ne 5$