If the circles x^{2}+y^{2}+2x+2ky+6=0 and x^{ | vedclass

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(A) Two circles $x^{2}+y^{2}+2g_{1}x+2f_{1}y+c_{1}=0$ and $x^{2}+y^{2}+2g_{2}x+2f_{2}y+c_{2}=0$ intersect orthogonally if and only if $2(g_{1}g_{2}+f_

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

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(A) Two circles $x^{2}+y^{2}+2g_{1}x+2f_{1}y+c_{1}=0$ and $x^{2}+y^{2}+2g_{2}x+2f_{2}y+c_{2}=0$ intersect orthogonally if and only if $2(g_{1}g_{2}+f_{1}f_{2})=c_{1}+c_{2}$.For the given circles:Circle $1$: $g_{1}=1, f_{1}=k, c_{1}=6$Circle $2$: $g_{2}=0, f_{2}=k, c_{2}=k$Substituting these values into the condition:$2((1)(0) + (k)(k)) = 6 + k$$2k^{2} = 6 + k$$2k^{2} - k - 6 = 0$Factoring the quadratic equation:$2k^{2} - 4k + 3k - 6 = 0$$2k(k-2) + 3(k-2) = 0$$(k-2)(2k+3) = 0$Thus, $k = 2$ or $k = -\frac{3}{2}$.

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

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