1. Given parabola and point

The parabola is $y^2=4x$ which is of the form $y^2=4ax$ with $a=1$. So its:

• vertex is $(0,0)$
• focus is $(1,0)$
• axis is the $x$-axis.

The given point is $(a,0)$ with $a>0$.

We are told that the shortest distance from $(a,0)$ to the parabola is $4$.

2. Find shortest distance from $(a,0)$ to the parabola

Take a general point on the parabola: $P(t^2,2t)$ using the standard parametrization of $y^2=4x$.

Distance squared from $(a,0)$ to $P(t^2,2t)$ is $D^2=(t^2-a)^2+(2t)^2$ $=t^4-2at^2+a^2+4t^2$ $=t^4+(4-2a)t^2+a^2.$

To minimize $D$, minimize $D^2$.

Let $u=t^2 \ge 0$. Then $D^2=u^2+(4-2a)u+a^2.$

Differentiate with respect to $u$: $\frac{d}{du}D^2=2u+(4-2a).$

Set equal to zero: $2u+4-2a=0$ $u=a-2.$

So the minimum occurs at:

• $u=a-2$ if $a\ge 2$
• otherwise boundary point $u=0$.

Since shortest distance is given as $4$, let us use the minimum formula for $a\ge2$:

$D_{\min}^2=(a-2)^2+(4-2a)(a-2)+a^2.$

Simplify: $D_{\min}^2=a^2-4a+4+4a-8-2a^2+4a+a^2$ $=4a-4.$

Hence $D_{\min}=2\sqrt{a-1}.$

Given this equals $4$: $2\sqrt{a-1}=4$ $\sqrt{a-1}=2$ $a-1=4$ $a=5.$

So the point is $(5,0)$.

3. Find the required circle

The circle passes through:

• $(5,0)$
• focus $(1,0)$

Its centre lies on the axis of the parabola, i.e. on the $x$-axis.

Let the centre be $(h,0)$.

Since the circle passes through both $(1,0)$ and $(5,0)$, the distances from the centre to these points must be equal: $|h-1|=|h-5|.$

Thus $h$ is the midpoint of $1$ and $5$: $h=\frac{1+5}{2}=3.$

Radius: $r=|3-1|=2.$

Therefore the circle is $(x-3)^2+y^2=4.$

Expanding: $x^2-6x+9+y^2=4$ $x^2+y^2-6x+5=0.$

4. Check options

This matches:

Option B: $x^2+y^2-6x+5=0.$

5. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.