The equation of the normal to the circle $x^2+y^2+6 x+4 y-3=0$ at $(1,-2)$ is
The equation of the normal to the circle $x^2+y^2+6 x+4 y-3=0$ at $(1,-2)$ is
$y+1=0$
$y+2=0$
$y+3=0$
$y-2=0$
Solution
Given circle is
$x^2+y^2+6 x+4 y-3=0$
Equation of the tangent at $\left(x_1, y_1\right)$ is
$x x_1+y y_1+3\left(x+x_1\right)+2\left(y+y_1\right)-3=0$
If this line passes through $(1,-2)$, then
$\begin{array}{rrr} & x-2 y+3(x+1)+2(y-2)-3=0 \\ \Rightarrow & x-2 y+3 x+3+2 y-4-3=0 \\ \Rightarrow & 4 x-4=0 \\ \Rightarrow & x-1=0\end{array}$
Equation of the normal is
$0 \cdot x-1 \cdot y=\lambda$
It passes through $(1,-2)$
$\begin{aligned}-(-2) & =\lambda \Rightarrow \lambda=2 \\ \Rightarrow \quad-y & =2 \Rightarrow y+2=0\end{aligned}$