If the line through the point $P(5,3)$ meets the circle $x^2+y^2-2 x-4 y+\alpha=0$ at $\mathrm{A}(4,2)$ and…

If the line through the point $P(5,3)$ meets the circle $x^2+y^2-2 x-4 y+\alpha=0$ at $\mathrm{A}(4,2)$ and $\mathrm{B}\left(x_1, y_1\right)$, then $P A \cdot P B$ is equal to
  1. 6
  2. 12
  3. 9
  4. 8

Solution

Line through $P(5,3)$ meets the circle $x^2+y^2-2 x-4 y+\alpha=0 \text { at } A(4,2) \text { and } B\left(x_1, y_1\right)$ $\therefore A(4,2)$ lies on the circle $\Rightarrow 16+4-8-8+\alpha=0 \Rightarrow \alpha=-4$
Eq. of circle becomes $x^2+y^2-2 x-4 y-4=0$ Also, equation of line passing $P(5,3)$ and $A(4,2)$ is $(y-3)=\frac{2-3}{4-5}(x-5) \Rightarrow x=2+y$
Put in equation of circle $\begin{aligned} & (2+y)^2+y^2-2(2+y)-4 y-4=0 \\ & \Rightarrow y^2-y-2=0 \Rightarrow y=-1,2, \text { So, } x=1,4 \\ & \therefore B\left(x_1, y_1\right)=(1,-1) \\ & P A \cdot P B=\sqrt{(5-4)^2+(3-2)^2} \cdot \sqrt{(5-1)^2+(3+1)^2}=8 \end{aligned}$

Asked in: AP EAMCET 2024 (23 May Shift 1)

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