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
Step-by-Step Solution
Key Concept: The shortest distance from an external point to a parabola is along the normal; use the normal foot condition $a=2+t^2$ and the given distance $PR=4$ to find $t^2=3$, hence $a=5$; then write the circle with diameter joining $(5,0)$ and focus $(1,0)$.
Normal at $P=(t^2,2t)$ on $y^2=4x$: $y+tx=2t+t^3$. Substituting $(a,0)$: $a=2+t^2$, so the foot $R=(2+t^2,0)$. Shortest distance $PR=\sqrt{(t^2-a)^2+(2t)^2}=\sqrt{4t^2}=2|t|$. Wait — shortest distance from $(a,0)$ to parabola: $PR^2=(t^2-a)^2+4t^2=4t^2$ when foot is the normal foot, so $PR=2|t|$. But $PR=4$ gives $|t|=2$... Actually: $PR^2=(t^2-(2+t^2))^2+4t^2=4+4t^2$. Setting $PR=4$: $4+4t^2=16$, so $t^2=3$, $a=2+3=5$, $R=(5,0)$. Focus $S=(1,0)$. Circle through $(5,0)$ and $(1,0)$ with centre on $x$-axis: $(x-1)(x-5)+y^2=0$, i.e., $x^2+y^2-6x+5=0$.
Correct Answer: 2