Question

The circles $x^2+y^2+2kx+2y+6=0$ and $x^2+y^2+2kx+k=0$ intersect orthogonally when $k$ equals to

• A. $2$
• B. $\frac{1+\sqrt{193}}{16}$
• C. $\frac{1-\sqrt{193}}{16}$
• D. $-\frac{3}{2}$

Answer 1

Answer: (A), (D)

$-\frac{3}{2}$
$2$

Given circles are
$x^2+y^2+2kx+2y+6=0$ and $x^2+y^2+2kx+k=0$
Here, $g_1=-k,f_1=-1,c_1=6$
$g_2=-k,f_2=0,c_2=k$
Condition for othogonality
$2(g_1{g}_2+f_1{f}_2)=c_1+c_2$
$\therefore 2(k^2+0)=6+k$
$\therefore k=\frac{1\pm \sqrt{1+48}}{2\times 2}=\frac{1\pm 7}{4}$
$\Rightarrow k=2,-\frac{3}{2}$