If $2 x-3 y+3=0$ and $x+2 y+k=0$ are conjugate lines with respect to the circle $\mathrm{S} \equiv x^2+y^2+8…
If $2 x-3 y+3=0$ and $x+2 y+k=0$ are conjugate lines with respect to the circle $\mathrm{S} \equiv x^2+y^2+8 x-6 y-24=0$, then the length of the tangent drawn from the point $\left(\frac{k}{4}, \frac{k}{3}\right)$ to the circle $\mathrm{S}=0$ is
$7$
$1$
$12$
$24$
Solution
$\begin{aligned}& \text { Given equation of circle } x^2+y^2+8 x-6 y-24=0 \\& \Rightarrow(x+4)^2+(y-3)^2=24+25=49\end{aligned}$
So, centre $=(-4,3)$, radius $=7$
Since, $2 x-3 y+3=0$ and $x+2 y+k=0$ are conjugate lines with respect to the given circle.
So, $7^2(1 \times 2+2 \times(-3))=(2 \times(+4)+(-3) \times(-3)-3)$
$(1 \times(+4)+2 \times(-3)-k) \Rightarrow k=12$
Now, $\left(\frac{k}{4}, \frac{k}{3}\right)=(3,4)$ $\mathrm{AO}=\sqrt{(3+4)^2+(4-3)^2}=\sqrt{50}$
Since, In $\triangle \mathrm{APO} ; \mathrm{AP}=\sqrt{50-49}=1$
