Let the shortest distance from $(a,0)$, $a0$, 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 :

Question

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 :

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] For $(a,0)$ and a point $(x,2\sqrt x)$: $d^2=(x-a)^2+4x$ is minimised at $x=a-2$, giving $d^2=4a-4=16$, so $a=5$. The circle through $(5,0)$ and the focus $(1,0)$ with its centre on the $x$-axis has diameter from $1$ to $5$: centre $(3,0)$, radius $2$: $x^2+y^2-6x+5=0$. Answer (2).

Q. 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 :

Explanation

Reviewed by Dr. Abubacker

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