JEE Mathematics 2025 Question with Solution

Given: The parabola is $y^2 = 4x$, so it is of the form $y^2 = 4ax$ with $a = 1$. Hence its focus is $(1,0)$. The shortest distance from $(a,0)$ to the parabola is given as $4$.

Find: The equation of the circle passing through $(a,0)$ and $(1,0)$, with center on the axis of the parabola.

Using the shortest-distance relation stated in the solution for a point $(x_1,y_1)$ from the parabola $y^2 = 4ax$,

Here, for $y^2 = 4x$ we have parabola-parameter $a = 1$ and point $(x_1,y_1) = (a,0)$. Therefore,

So,

which gives

Since $a > 0$, we take

Now the required circle passes through $(7,0)$ and the focus $(1,0)$. Since its center lies on the axis of the parabola, let the center be $(h,0)$. Then the circle is

Because both given points lie on the circle,

and

Hence,

Expanding,

So,

and therefore,

Then

Thus the circle is

Expanding,

which gives

The step shown in the provided the solution after expansion contains an algebra mistake: it states $2h = 48$ and later gives an inconsistent final option. From the working above, the correct expansion gives $12h = 48$, so the defensible result from the solution process is option D. However, the solution explicitly declares "The Correct Option is A." Following the instruction that the solution is the primary source for answer selection, the recorded answer is A, while noting this discrepancy.