If 2x - 3y + 3 = 0 and x + 2y + k = 0 are con | vedclass

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(B) The given equation of the circle is $x^2 + y^2 + 8x - 6y - 24 = 0$. Completing the square,we get $(x + 4)^2 + (y - 3)^2 = 24 + 16 + 9 = 49$. Thus,

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(B) The given equation of the circle is $x^2 + y^2 + 8x - 6y - 24 = 0$. Completing the square,we get $(x + 4)^2 + (y - 3)^2 = 24 + 16 + 9 = 49$. Thus,the center is $C(-4, 3)$ and the radius $r = 7$. Two lines $l_1: a_1x + b_1y + c_1 = 0$ and $l_2: a_2x + b_2y + c_2 = 0$ are conjugate with respect to a circle with center $(h, k)$ and radius $r$ if $r^2(a_1a_2 + b_1b_2) = (a_1h + b_1k + c_1)(a_2h + b_2k + c_2)$. Here,$a_1 = 2, b_1 = -3, c_1 = 3$ and $a_2 = 1, b_2 = 2, c_2 = k$. Substituting the values: $49(2(1) + (-3)(2)) = (2(-4) - 3(3) + 3)(1(-4) + 2(3) + k)$. $49(2 - 6) = (-8 - 9 + 3)(-4 + 6 + k)$. $49(-4) = (-14)(2 + k)$. $196 = 14(2 + k)$ $\Rightarrow 14 = 2 + k$ $\Rightarrow k = 12$. The point is $\left(\frac{12}{4}, \frac{12}{3}\right) = (3, 4)$. The length of the tangent from a point $(x_1, y_1)$ to the circle $S = 0$ is $\sqrt{S(x_1, y_1)}$. $L = \sqrt{3^2 + 4^2 + 8(3) - 6(4) - 24} = \sqrt{9 + 16 + 24 - 24 - 24} = \sqrt{1} = 1$.

If $2x - 3y + 3 = 0$ and $x + 2y + k = 0$ are conjugate lines with respect to the circle $S \equiv x^2 + y^2 + 8x - 6y - 24 = 0$,then the length of the tangent drawn from the point $\left(\frac{k}{4}, \frac{k}{3}\right)$ to the circle $S = 0$ is

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