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

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(A) The condition for two circles $x^{2}+y^{2}+2g_{1}x+2f_{1}y+c_{1}=0$ and $x^{2}+y^{2}+2g_{2}x+2f_{2}y+c_{2}=0$ to intersect orthogonally is $2g_{1}

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$2$ or $-3/2$

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(A) The condition for two circles $x^{2}+y^{2}+2g_{1}x+2f_{1}y+c_{1}=0$ and $x^{2}+y^{2}+2g_{2}x+2f_{2}y+c_{2}=0$ to intersect orthogonally is $2g_{1}g_{2}+2f_{1}f_{2}=c_{1}+c_{2}$.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) + 2(k)(k) = 6 + k$$0 + 2k^{2} = 6 + k$$2k^{2} - k - 6 = 0$Factoring the quadratic equation:$(2k+3)(k-2) = 0$Thus,$k = 2$ or $k = -3/2$.

If the circles $x^{2}+y^{2}+2x+2ky+6=0$ and $x^{2}+y^{2}+2ky+k=0$ intersect orthogonally,then $k$ is

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