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If the circles $x^{2}+y^{2}+2 x+2 k y+6=0$ and $x^{2}+y^{2}+2 k y+k=0$ intersect orthogonally. then $k$ is equal to
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The correct answer is:
2 or $-\frac{3}{2}$
Two circles are orthogonally if and only if $\quad 2\left(g_{1} g_{2}+f_{1} f_{2}\right)=c_{1}+c_{2}$
$\Rightarrow \quad 2[(1 \times 0+(k) k]=6+k$
$\Rightarrow \quad 2 k^{2}=6+k$
$\Rightarrow \quad 2 k^{2}-k-6=0$
$\Rightarrow \quad 2 k^{2}-4 k+3 k-6=0$
$\Rightarrow \quad \quad 2 k(k-2)+3(k-2)=0$
$\Rightarrow \quad(k-2)(2 k+3)=0$
$\Rightarrow \quad k=2-\frac{3}{2}$
$\Rightarrow \quad 2[(1 \times 0+(k) k]=6+k$
$\Rightarrow \quad 2 k^{2}=6+k$
$\Rightarrow \quad 2 k^{2}-k-6=0$
$\Rightarrow \quad 2 k^{2}-4 k+3 k-6=0$
$\Rightarrow \quad \quad 2 k(k-2)+3(k-2)=0$
$\Rightarrow \quad(k-2)(2 k+3)=0$
$\Rightarrow \quad k=2-\frac{3}{2}$
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