Let circle C be the image of x^2+y^2-2x+4y-4= | vedclass

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(D) The given circle is $x^2+y^2-2x+4y-4=0$. Its centre is $(1, -2)$ and radius $r = \sqrt{1^2 + (-2)^2 - (-4)} = \sqrt{1+4+4} = 3$.Let the centre of

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$4$

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(D) The given circle is $x^2+y^2-2x+4y-4=0$. Its centre is $(1, -2)$ and radius $r = \sqrt{1^2 + (-2)^2 - (-4)} = \sqrt{1+4+4} = 3$.Let the centre of circle $C$ be $O(h, k)$. The reflection of $(1, -2)$ about the line $2x-3y+5=0$ is given by:$\frac{h-1}{2} = \frac{k+2}{-3} = \frac{-2(2(1)-3(-2)+5)}{2^2+(-3)^2} = \frac{-2(2+6+5)}{13} = -2$.Thus,$h-1 = -4 \Rightarrow h = -3$ and $k+2 = 6 \Rightarrow k = 4$. So,$O = (-3, 4)$.The equation of circle $C$ is $(x+3)^2+(y-4)^2 = 3^2 = 9$.Point $A$ lies on $C$ such that $OA$ is parallel to the $x$-axis and $A$ is to the right of $O$. Since $O=(-3, 4)$ and $r=3$,$A = (-3+3, 4) = (0, 4)$.The length of arc $AB$ is $1/6$ of the perimeter,so the central angle $	heta = \frac{1}{6} 	imes 2\pi = \frac{\pi}{3}$.Since $B(\alpha, \beta)$ lies on $C$ and $\beta Then $\beta - \sqrt{3}\alpha = (4 - 3\sqrt{3}/2) - \sqrt{3}(-1.5) = 4 - 1.5\sqrt{3} + 1.5\sqrt{3} = 4$.

List-$I$	List-$II$
$A$. The equation of the polar of $(4, 3)$ with respect to $C$	$I$. $y+5=0$
$B$. The equation of the tangent at $(9, -5)$ on $C$	$II$. $x=1$
$C$. The equation of the normal at $(-7, -5)$ on $C$	$III$. $3x+8y=27$
$D$. The equation of the diameter passing through $(1, -5)$ and $(1, 3)$	$IV$. $x=9$

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