If the length of the chord $2 x+3 y+k=0$ of the circle $x^2+y^2-6 x-8 y+9=0$ is $2 \sqrt{3}$, then one of…
- 31
- 5
- $-5$
- $-13$
Solution

$\Rightarrow \quad A M=\frac{A B}{2}=\frac{2 \sqrt{3}}{2}=\sqrt{3}$ And $C M=\frac{2 \times 3+4 \times 3+k}{\sqrt{4+9}}=\frac{18+k}{\sqrt{13}}$ In $\triangle A C M, A C^2=C M^2+A M^2$ $\begin{aligned} & \Rightarrow \quad 4^2=\left(\frac{18+k}{\sqrt{13}}\right)^2+(\sqrt{3})^2 \\ & \Rightarrow \quad 16-3=\left(\frac{18+k}{\sqrt{13}}\right)^2 \\ & \therefore \quad \frac{18+k}{\sqrt{13}}=\sqrt{13} \Rightarrow 18+k=13 \Rightarrow k=-5\end{aligned}$
Asked in: AP EAMCET 2023 (16 May Shift 1)