The equation of a line parallel to the line $3 x+4 y=0$ and touching the circle $x^{2}+y^{2}=9$ in the first quadrant, is

Equation of circle is $x^{2}+y^{2}=9$ ...(i) And equation of line is $3 x+4 y=0$ ....(ii) Equation of line parallel to the line (ii) . $=$ $$ \begin{aligned} 3 x+4 y &=k \\ y &=\frac{-3 x}{4}+\frac{k}{4} ...(iii) \end{aligned} $$ $\because$ Circle touches the line (iii), so by the condition of tangency $$ c^{2}=a^{2}\left(1+m^{2}\right) $$ $\Rightarrow \quad\left(\frac{k}{4}\right)^{2}=9\left[1+\left(\frac{-3}{4}\right)^{2}\right]$ $\left[\because c=\frac{k}{4}, a=3, m=\frac{-3}{4}\right]$ $\Rightarrow$ $$ k=\pm 15 $$ Hence, $3 x+4 y=15$ touches the circle in the first quadrant.