Consider the point $P(\alpha, \beta)$ on the line $2 x+y=1$. If the $P$ and $(3,2)$ are conjugate points…

Consider the point $P(\alpha, \beta)$ on the line $2 x+y=1$. If the $P$ and $(3,2)$ are conjugate points with respect to the circle $x^2+y^2=4$, then $\alpha+\beta=$
  1. 3
  2. -1
  3. -5
  4. 7

Solution

Polar of point $(3,2)$ w.r.t. circle $x^2+y^2=4 \text { is } T_1=0 \Rightarrow 3 x+2 y=4$
Since, $(3,2)$ and $P(\alpha, \beta)$ are conjugate point. So, polar of $(3,2)$ passes through $P(\alpha, \beta) \Rightarrow 3 \alpha+2 \beta=4$ ...(i) Also, $P(\alpha, \beta)$ lies on $2 x+y=1 \Rightarrow 2 \alpha+\beta=1$ ...(ii) Solving (i) and (ii), $\alpha=-2, \beta=5$ $\therefore \alpha+\beta=-2+5=3 \text {. }$

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

Practice more Circle questions on Aicharya