Question

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