Q93 / 160 · AP EAMCET 2023 (16 May Shift 1) · Mathematics
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 the values of $\mathrm{k}$ is
- A31
- B5
- Correct answerC$-5$
- D$-13$
Solution
The given equation of circle is
$x^2+y^2-6 x-8 y+9=0$
Its centre is $C=(3,4)$
and Radius $R=\sqrt{9+16-9} \Rightarrow R=\sqrt{16}=4$
$\because A B=2 \sqrt{3}$
$\Rightarrow \quad A M=\frac{A B}{2}=\frac{2 \sqrt{3}}{2}=\sqrt{3}$ And $C M=\frac{2 \times 3+4 \times 3+k}{\sqrt{4+9}}=\frac{18+k}{\sqrt{13}}$ In $\triangle A C M, A C^2=C M^2+A M^2$ $\begin{aligned} & \Rightarrow \quad 4^2=\left(\frac{18+k}{\sqrt{13}}\right)^2+(\sqrt{3})^2 \\ & \Rightarrow \quad 16-3=\left(\frac{18+k}{\sqrt{13}}\right)^2 \\ & \therefore \quad \frac{18+k}{\sqrt{13}}=\sqrt{13} \Rightarrow 18+k=13 \Rightarrow k=-5\end{aligned}$
$\Rightarrow \quad A M=\frac{A B}{2}=\frac{2 \sqrt{3}}{2}=\sqrt{3}$ And $C M=\frac{2 \times 3+4 \times 3+k}{\sqrt{4+9}}=\frac{18+k}{\sqrt{13}}$ In $\triangle A C M, A C^2=C M^2+A M^2$ $\begin{aligned} & \Rightarrow \quad 4^2=\left(\frac{18+k}{\sqrt{13}}\right)^2+(\sqrt{3})^2 \\ & \Rightarrow \quad 16-3=\left(\frac{18+k}{\sqrt{13}}\right)^2 \\ & \therefore \quad \frac{18+k}{\sqrt{13}}=\sqrt{13} \Rightarrow 18+k=13 \Rightarrow k=-5\end{aligned}$
Hence, the correct answer is option: 'C'.