Consider the point P(alpha, beta) on the line | vedclass

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(A) The equation of the polar of point $(3,2)$ with respect to the circle $x^2+y^2=4$ is given by $T=0$,which is $3x+2y=4$. Since $P(\alpha, \beta)$ a

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$3$

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(A) The equation of the polar of point $(3,2)$ with respect to the circle $x^2+y^2=4$ is given by $T=0$,which is $3x+2y=4$. Since $P(\alpha, \beta)$ and $(3,2)$ are conjugate points,the polar of $(3,2)$ must pass through $P(\alpha, \beta)$. Thus,$3\alpha+2\beta=4$ ...$(i)$. Given that $P(\alpha, \beta)$ lies on the line $2x+y=1$,we have $2\alpha+\beta=1$ ...(ii). Solving equations $(i)$ and (ii): From (ii),$\beta = 1-2\alpha$. Substituting into $(i)$: $3\alpha + 2(1-2\alpha) = 4$ $\Rightarrow 3\alpha + 2 - 4\alpha = 4$ $\Rightarrow -\alpha = 2$ $\Rightarrow \alpha = -2$. Then $\beta = 1 - 2(-2) = 5$. Therefore,$\alpha+\beta = -2+5 = 3$.

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

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