If the circles x^2+y^2+kx+4y+2=0 and 2(x^2+y^ | vedclass

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(A) The given equations of the circles are:$S_1: x^2+y^2+kx+4y+2=0$$S_2: x^2+y^2-2x-\frac{3}{2}y+\frac{k}{2}=0$For two circles $x^2+y^2+2g_1x+2f_1y+c_

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$\frac{-10}{3}$

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(A) The given equations of the circles are:$S_1: x^2+y^2+kx+4y+2=0$$S_2: x^2+y^2-2x-\frac{3}{2}y+\frac{k}{2}=0$For two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ to cut orthogonally,the condition is $2g_1g_2+2f_1f_2=c_1+c_2$.Here,$g_1 = \frac{k}{2}, f_1 = 2, c_1 = 2$ and $g_2 = -1, f_2 = -\frac{3}{4}, c_2 = \frac{k}{2}$.Substituting these values into the condition:$2(\frac{k}{2})(-1) + 2(2)(-\frac{3}{4}) = 2 + \frac{k}{2}$$-k - 3 = 2 + \frac{k}{2}$$-5 = k + \frac{k}{2}$$-5 = \frac{3k}{2}$$k = -\frac{10}{3}$.

If the circles $x^2+y^2+kx+4y+2=0$ and $2(x^2+y^2)-4x-3y+k=0$ cut orthogonally,then $k=$

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