jee-main 2025 Q1

jee-main · India · session1_24jan_shift1 Circles Circle Equation Derivation
Let circle $C$ be the image of $x^2 + y^2 - 2x + 4y - 4 = 0$ in the line $2x - 3y + 5 = 0$ and $A$ be the point on $C$ such that $OA$ is parallel to $x$-axis and $A$ lies on the right hand side of the centre $O$ of $C$. If $B(\alpha, \beta)$, with $\beta < 4$, lies on $C$ such that the length of the arc $AB$ is $(1/6)^{\text{th}}$ of the perimeter of $C$, then $\beta - \sqrt{3}\alpha$ is equal to
(1) $3 + \sqrt{3}$
(2) 4
(3) $4 - \sqrt{3}$
(4) 3
Let circle $C$ be the image of $x^2 + y^2 - 2x + 4y - 4 = 0$ in the line $2x - 3y + 5 = 0$ and $A$ be the point on $C$ such that $OA$ is parallel to $x$-axis and $A$ lies on the right hand side of the centre $O$ of $C$. If $B(\alpha, \beta)$, with $\beta < 4$, lies on $C$ such that the length of the arc $AB$ is $(1/6)^{\text{th}}$ of the perimeter of $C$, then $\beta - \sqrt{3}\alpha$ is equal to\\
(1) $3 + \sqrt{3}$\\
(2) 4\\
(3) $4 - \sqrt{3}$\\
(4) 3