If \(\quad 2 x^{2}+2 y^{2}+4 x+5 y+1=0 \quad\) and \(3 x^{2}+3 y^{2}+6 x-7 y+3 k=0 \quad\) are orthogonal, then value of \(k\) is

Given equation of circles
\[
\begin{aligned}
2 x^{2}+2 y^{2}+4 x+5 y+1 &=0 \\
\Rightarrow \quad x^{2}+y^{2}+2 x+5 / 2 y+1 / 2 &=0 \quad \ldots \text { (i) } \\
3 x^{2}+3 y^{2}+6 x-7 y+3 k &=0 \\
\Rightarrow \quad x^{2}+y^{2}+2 x+7 / 3+k &=0 \quad \ldots \text { (ii) }
\end{aligned}
\]
Centre and radius of first circle and second circle
\[
\begin{gathered}
C_{1} \rightarrow(-1,-5 / 4), \\
R_{1} \rightarrow \sqrt{1+\frac{25}{16}-1 / 2}=\frac{\sqrt{16+25-8}}{16}=\sqrt{\frac{33}{16}} \\
C_{2} \rightarrow(-1,7 / 6), \\
R_{2} \rightarrow \sqrt{1+\frac{49}{36}-k}=\sqrt{\frac{85-36 k}{36}}
\end{gathered}
\]
Now, condition for orthogonality of two circles
\[
\begin{aligned}
\left(C_{1} C_{2}\right)^{2} &=R_{1}{ }^{2}+R_{2}{ }^{2} \\
(-1+1)^{2}+(7 / 6+5 / 4)^{2} &=\frac{33}{16}+\frac{85-36 k}{36} \\
0+\left(\frac{29}{12}\right)^{2} &=\frac{33}{16}+\frac{85-36 k}{36}
\end{aligned}
\]
\(\begin{aligned} \frac{841}{144} &=\frac{33}{16}+\frac{85-36 k}{36} \\ 841 &=297+340-144 k \\ 144 k &=637-841=-204 \\ k &=-\frac{17}{12} \end{aligned}\)