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
$2\left(g_{1} g_{2}+f_{1} f_{2}\right)=c_{1}+c_{2}$
$\Rightarrow 2[(1 \times 0+(k) k]=6 +k$
$\Rightarrow 2 k^{2}=6+k$
$\Rightarrow 2 k^{2}-k-6=0$
$\Rightarrow 2 k^{2}-4 k+3 k-6=0$
$\Rightarrow 2 k(k-2)+3(k-2)=0$
$\Rightarrow (k-2)(2 k+3)=0$
$\Rightarrow k=2,-\frac{3}{2}$