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(C) Given circles are $S_1: x^2 + y^2 + kx - 4y - 1 = 0$ and $S_2: x^2 + y^2 - \frac{14}{3}x + \frac{23}{3}y - 5 = 0$.Since $S_1$ and $S_2$ are orthog

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$5x^2 + 5y^2 - 22x + 15y - 17 = 0$

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(C) Given circles are $S_1: x^2 + y^2 + kx - 4y - 1 = 0$ and $S_2: x^2 + y^2 - \frac{14}{3}x + \frac{23}{3}y - 5 = 0$.Since $S_1$ and $S_2$ are orthogonal,the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$ must hold.Here,$g_1 = \frac{k}{2}, f_1 = -2, c_1 = -1$ and $g_2 = -\frac{7}{3}, f_2 = \frac{23}{6}, c_2 = -5$.Substituting these values: $2(\frac{k}{2})(-\frac{7}{3}) + 2(-2)(\frac{23}{6}) = -1 - 5$.$-\frac{7k}{3} - \frac{46}{3} = -6$ $\Rightarrow -7k - 46 = -18$ $\Rightarrow -7k = 28$ $\Rightarrow k = -4$.The family of circles passing through the intersection of $S_1$ and $S_2$ is $S_1 + \lambda(S_2') = 0$,where $S_2'$ is the normalized form of $S_2$.$x^2 + y^2 - 4x - 4y - 1 + \lambda(x^2 + y^2 - \frac{14}{3}x + \frac{23}{3}y - 5) = 0$.Passing through $(-1, -1)$: $(1 + 1 + 4 + 4 - 1) + \lambda(1 + 1 + \frac{14}{3} - \frac{23}{3} - 5) = 0$.$9 + \lambda(2 - 3 - 5) = 0$ $\Rightarrow 9 - 6\lambda = 0$ $\Rightarrow \lambda = \frac{3}{2}$.Substituting $\lambda = \frac{3}{2}$ into the equation: $x^2 + y^2 - 4x - 4y - 1 + \frac{3}{2}(x^2 + y^2 - \frac{14}{3}x + \frac{23}{3}y - 5) = 0$.Multiplying by $2$: $2x^2 + 2y^2 - 8x - 8y - 2 + 3x^2 + 3y^2 - 14x + 23y - 15 = 0$.$5x^2 + 5y^2 - 22x + 15y - 17 = 0$.

The equation of the circle passing through the points of intersection of the two orthogonal circles $S_1 = x^2 + y^2 + kx - 4y - 1 = 0$ and $S_2 = 3x^2 + 3y^2 - 14x + 23y - 15 = 0$ and passing through the point $(-1, -1)$ is:

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