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 equal to

• 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

Answer: (A)

Two circles are orthogonally if and only if
$2\left(g_1{g}_2+f_1{f}_2\right)=c_1+c_2$
$\Rightarrow 2[(1\times 0+(k)k]=6+k$
$\Rightarrow 2k^2=6+k$
$\Rightarrow 2k^2-k-6=0$
$\Rightarrow 2k^2-4k+3k-6=0$
$\Rightarrow 2k(k-2)+3(k-2)=0$
$\Rightarrow (k-2)(2k+3)=0$
$\Rightarrow k=2,-\frac{3}{2}$