Question

If the circle $x^2+y^2+2x+2ky+6=0,$ $x^2+y^2+2ky+k=0$ intersect orthogonally, then $k$ is

• A. $2$ or $-\frac{3}{2}$
• B. $-2$ or $-\frac{3}{2}$
• C. $2$ or $\frac{3}{2}$
• D. $-2$ or $\frac{3}{2}$

Answer 1

The correct option is A $2$ or $-\frac{3}{2}$

Since, circle cut orthogonally, if
$2g{g}^{\prime }+2f{f}^{\prime }=c+c^{\prime }$
$\Rightarrow 0+2k^2=6+k$
$\Rightarrow 2k^2-k-6=0$
$\Rightarrow (2k+3)(k-2)=0$
$\Rightarrow k=2or-3/2$