If the angle between the circles x^2+y^2-4x-6 | vedclass

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(C) Two circles are orthogonal if the angle between them is $\frac{\pi}{2}$.For circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$,th

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$-15$

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(C) Two circles are orthogonal if the angle between them is $\frac{\pi}{2}$.For circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$,the condition for orthogonality is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.Comparing the given equations with the standard form:Circle $1$: $g_1 = -2, f_1 = -3, c_1 = k$.Circle $2$: $g_2 = 4, f_2 = -2, c_2 = 11$.Substituting these values into the condition:$2(-2)(4) + 2(-3)(-2) = k + 11$$-16 + 12 = k + 11$$-4 = k + 11$$k = -15$.

If the angle between the circles $x^2+y^2-4x-6y+k=0$ and $x^2+y^2+8x-4y+11=0$ is $\frac{\pi}{2}$,then the value of $k$ is

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