The equation of a tangent to the circle $x^2+y^2+2 x-12 y$ $-132=0$ which is perpendicular to the line $12…
The equation of a tangent to the circle $x^2+y^2+2 x-12 y$ $-132=0$ which is perpendicular to the line $12 x+5 y+k=0$ is
$5 x-12 y+92=0$
$5 \mathrm{x}-12 \mathrm{y}-246=0$
$5 \mathrm{x}-12 \mathrm{y}-169=0$
$5 x-12 y+246=0$
Solution
Given equation of circle
$\begin{aligned}
& x^2+y^2+2 x-12 y-132=0 \\
& \Rightarrow(x+1)^2+(y-6)^2=13^2 \\
& \Rightarrow \text { Radius }=13, \text { Centre }=(-1,6)
\end{aligned}$
Since slope of the line $12 x+5 y+8=0$ is
$\mathrm{m}_1=\frac{-12}{5}$
So slope of perpendicular line to the given line is
$\mathrm{m}_2=\frac{5}{12}$
Now, equation of required line is $y=\frac{5}{12} x+C$ Since, perpendicular distance of tangent from centre $=13$
$\begin{aligned}
& \Rightarrow \frac{\left|\frac{5}{12} \times(-1)+C-6\right|}{\sqrt{\left(\frac{5}{12}\right)^2+1}}=13 \\
& \Rightarrow|-77+12 C|=169 \Rightarrow C=-\frac{92}{12} \text { or } \frac{246}{12}
\end{aligned}$
so, $y=\frac{5 x}{12}-\frac{92}{12} \Rightarrow 5 x-12 y-92=0$
$\text { or, } y=\frac{5}{12} x+\frac{246}{12} \Rightarrow 5 x-12 y+246=0$