If the lines 2x + y + 12 = 0 and kx - 3y - 10 | vedclass

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(A) The condition for two lines $L_1: a_1x + b_1y + c_1 = 0$ and $L_2: a_2x + b_2y + c_2 = 0$ to be conjugate with respect to the circle $x^2 + y^2 +

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$4$

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(A) The condition for two lines $L_1: a_1x + b_1y + c_1 = 0$ and $L_2: a_2x + b_2y + c_2 = 0$ to be conjugate with respect to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $r^2(a_1a_2 + b_1b_2) = (g a_1 + f b_1 - c_1)(g a_2 + f b_2 - c_2)$.For the circle $x^2 + y^2 - 4x + 3y - 1 = 0$,we have $g = -2$,$f = 3/2$,and $c = -1$.The radius squared is $r^2 = g^2 + f^2 - c = (-2)^2 + (3/2)^2 - (-1) = 4 + 9/4 + 1 = 29/4$.For line $L_1: 2x + y + 12 = 0$,$a_1 = 2, b_1 = 1, c_1 = 12$.For line $L_2: kx - 3y - 10 = 0$,$a_2 = k, b_2 = -3, c_2 = -10$.Substitute these into the condition:$\frac{29}{4}(2k - 3) = (-2(2) + \frac{3}{2}(1) - 12)(-2(k) + \frac{3}{2}(-3) - (-10))$$\frac{29}{4}(2k - 3) = (-4 + 1.5 - 12)(-2k - 4.5 + 10)$$\frac{29}{4}(2k - 3) = (-14.5)(-2k + 5.5)$$\frac{29}{4}(2k - 3) = -\frac{29}{2}(-2k + 5.5)$Divide both sides by $29/2$:$\frac{1}{2}(2k - 3) = -(-2k + 5.5)$$k - 1.5 = 2k - 5.5$$k = 4$.

If the lines $2x + y + 12 = 0$ and $kx - 3y - 10 = 0$ are conjugate with respect to the circle $x^2 + y^2 - 4x + 3y - 1 = 0$,then $k =$

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