The normal drawn at $(1,1)$ to the circle $x^2+y^2-4 x$ $+6 y-4=0$ is

The normal drawn at $(1,1)$ to the circle $x^2+y^2-4 x$ $+6 y-4=0$ is
  1. $4 x+3 y=7$
  2. $4 x+y=5$
  3. $x+y=2$
  4. $4 x-y=3$

Solution

Given circle is $x^2+y^2-4 x+6 y-4=0$. $\Rightarrow(g, f)=(-2,3)$
Equation of normal at $\left(x_1, y_1\right)$ is $\begin{aligned} & \left(y-y_1\right)\left(\frac{y_1+f}{x_1+g}\right)\left(x-x_1\right) \Rightarrow(y-1)=\left(\frac{1+3}{1-2}\right)(x-1) \\ & \Rightarrow 4 x+y=5 \end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 2)

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