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(A) The parabola is $y^2 = 4x$,so its focus is $S(1, 0)$.Let $P(t^2, 2t)$ be a point on the parabola. The normal at $P$ is $y + tx = 2t + t^3$.Since t

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$x^2+y^2-6x+5=0$

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(A) The parabola is $y^2 = 4x$,so its focus is $S(1, 0)$.Let $P(t^2, 2t)$ be a point on the parabola. The normal at $P$ is $y + tx = 2t + t^3$.Since the shortest distance from $(a, 0)$ to the parabola is along the normal,the normal passes through $(a, 0)$.Thus,$0 + t(a) = 2t + t^3$,which gives $a = 2 + t^2$ (since $t 
eq 0$ for shortest distance).The distance $PR = 4$,where $R = (a, 0) = (2+t^2, 0)$.$PR^2 = (t^2+t^2-2-t^2)^2 + (2t-0)^2 = (t^2-2)^2 + 4t^2 = t^4 - 4t^2 + 4 + 4t^2 = t^4 + 4$.Given $PR = 4$,so $PR^2 = 16$.$t^4 + 4 = 16$ $\Rightarrow t^4 = 12$ $\Rightarrow t^2 = \sqrt{12} = 2\sqrt{3}$.Wait,re-evaluating: The distance from $(a, 0)$ to $(t^2, 2t)$ is $\sqrt{(t^2-a)^2 + (2t)^2} = 4$.Since $a = 2+t^2$,$(t^2 - (2+t^2))^2 + 4t^2 = 16$ $\Rightarrow (-2)^2 + 4t^2 = 16$ $\Rightarrow 4 + 4t^2 = 16$ $\Rightarrow 4t^2 = 12$ $\Rightarrow t^2 = 3$.Then $a = 2 + 3 = 5$. The point is $(5, 0)$.The circle passes through $(5, 0)$ and the focus $(1, 0)$,and its center lies on the $x$-axis.The diameter of the circle is the segment joining $(1, 0)$ and $(5, 0)$.The equation is $(x-1)(x-5) + y^2 = 0$.$x^2 - 6x + 5 + y^2 = 0 \Rightarrow x^2 + y^2 - 6x + 5 = 0$.

Let the shortest distance from $(a, 0)$,$a > 0$,to the parabola $y^2 = 4x$ be $4$. Then the equation of the circle passing through the point $(a, 0)$ and the focus of the parabola,and having its centre on the axis of the parabola is:

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