Circle

$S \equiv x^2+y^2-6 x+8 y+k=0$$\Rightarrow (x-3)^2+(y+4)^2=25-k$Thus, the center of the circle $C$ is at $(3,-4)= (x_1, y_1)$.Let point $P$ as $(x_2, y_2)$ and point $Q$ is the point that divides the line segment $C P$ in the ratio $-1: 2$Also, $3 x+4 y-43=0$$\Rightarrow y=\frac{-3}{4}x+\frac{43}{3}$So, slope of tangent line $=\frac{-3}{4}$And slope of line $C P=-\left(\frac{-4}{3}\right)=\frac{4}{3}$Thus, equation of line $C P$ is$\Rightarrow y-(-4)=\frac{4}{3}(x-3)$$\Rightarrow y+4=\frac{4}{3}(x-3)$$\Rightarrow y=\frac{4}{3}x-8$Since, $P$ lies on both the lines $3 x+4 y-43=0$ and the line $C P$,$\therefore 3x+4\left(\frac{4}{3}x-8\right)-43=0$$\Rightarrow 3x+\frac{16}{3}x-32-43=0$$\Rightarrow \frac{25}{3}x=75$$\Rightarrow x=9$Now, $y=\frac{4}{3}x-8=\frac{4}{3}(9)-8$$=12-8=4$So, $P=(9,4)$Using section formula, we find the coordinates of $Q(x, y)$.$x=\frac{2x_1-x_2}{2-1}=\frac{2(3)-9}{1}=-3$$y=\frac{2y_1-y_2}{2-1}=\frac{2(-4)-4}{1}=-12$So, $Q=(x, y)=(-3,-12)$The power of point $Q$ w.r.t to circle, $x^2+y^2-6x+8y+k=0$ is$(-3)^2+(-12)^2-6(-3)+8(-12)+k$$=9+144+18-96+k$$=75+k \cdot\cdot\cdot\cdot\cdot\cdot\cdot (i)$Now, the line $3x+4y-43=0$ is tangent to the circle, the radius of the circle is the distance from the center $C(3,-4)$ to the line $3x+4y-43=0$So, $r=\frac{\left| A x_0+B y_0+C \right|}{\sqrt{A^2+B^2}}$$=\frac{|3(3)+4(-4)-43|}{\sqrt{3^2+4^2}}$$=\frac{|9-16-43|}{5}=\frac{50}{5}=10$Now, $r=\sqrt{25-k}$$\Rightarrow 10=\sqrt{25-k} \Rightarrow 25-k=100$$\Rightarrow k=-75$Put $k=-75$ in Eq. (i), we get The power of $Q$ w.r.t the circle$=75+k=75-75=0$