The vertex of the parabola (y - 1)^2 = 8(x - | vedclass

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(A) The given parabola is $(y - 1)^2 = 8(x - 1)$.Comparing with $(y - k)^2 = 4a(x - h)$,the vertex is $(h, k) = (1, 1)$.Since the vertex is the center

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$x^2 + y^2 - 2x - 2y - 18 = 0$

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(A) The given parabola is $(y - 1)^2 = 8(x - 1)$.Comparing with $(y - k)^2 = 4a(x - h)$,the vertex is $(h, k) = (1, 1)$.Since the vertex is the center of the circle,the equation of the circle is $(x - 1)^2 + (y - 1)^2 = r^2$.The latus rectum of the parabola is at $x - 1 = a$,where $4a = 8$,so $a = 2$. Thus $x = 3$.Substituting $x = 3$ into the parabola equation: $(y - 1)^2 = 8(3 - 1) = 16$,so $y - 1 = \pm 4$,giving $y = 5$ or $y = -3$.The ends of the latus rectum are $(3, 5)$ and $(3, -3)$.Since these points lie on the circle,they satisfy $(x - 1)^2 + (y - 1)^2 = r^2$.For $(3, 5)$: $(3 - 1)^2 + (5 - 1)^2 = 2^2 + 4^2 = 4 + 16 = 20 = r^2$.The equation of the circle is $(x - 1)^2 + (y - 1)^2 = 20$.Expanding this: $x^2 - 2x + 1 + y^2 - 2y + 1 = 20$.$x^2 + y^2 - 2x - 2y - 18 = 0$.

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 $(1, 3)$	$IV$. $x=9$

The vertex of the parabola $(y - 1)^2 = 8(x - 1)$ is at the centre of a circle and the parabola cuts that circle at the ends of its latus rectum. Then the equation of that circle is

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