Question

If the circles $x^2+y^2+2x+2ky+6=0$ and $x^2+y^2+2ky+k=0$ 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 $-\frac{3}{2}$

Given two circles are $x^2+y^2+2x+2ky+6=0$ and $x^2+y^2+2ky+k=0$

Using condition of orthogonality,

$2f_1{f}_2+2g_1{g}_2=c_1+c_2$

$\Rightarrow 2k^2+0=6+k\Rightarrow 2k^2-k-6=0$

$\Rightarrow (k-2)(2k+3)=0\Rightarrow k=2,-\frac{3}{2}$