Let the shortest distance from the point (a,0), a0, to the parabola y^2=4x be 4.

Question

Let the shortest distance from the point $(a,0)$, $a>0$, 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:

Accepted Answer

Normal to $y^2=4x$ at parameter $t$ meets the $x$-axis at $(2+t^2,0)$. Distance from $P=(t^2,2t)$ to this point:

$$PR^2=(t^2-(2+t^2))^2+4t^2=4+4t^2=16 \;\Rightarrow\; t^2=3,\quad a=2+3=5.$$

So $(a,0)=(5,0)$. The circle through $(5,0)$ and the focus $(1,0)$ with centre on the $x$-axis has these as diameter endpoints: centre $(3,0)$, radius $2$.

$$(x-3)^2+y^2=4 \;\Rightarrow\; x^2+y^2-6x+5=0.$$

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