JEE MAIN - Mathematics (2025 - 23rd January Evening Shift - No. 19)
Let the shortest distance from $(a, 0), a>0$, to the parabola $y^2=4 x$ 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 :
$x^2+y^2-8 x+7=0$
$x^2+y^2-6 x+5=0$
$x^2+y^2-4 x+3=0$
$x^2+y^2-10 x+9=0$
Explanation
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$$\begin{aligned} &\text { Normal at } \mathrm{P}\\ &\begin{gathered} y+t x=2 t+t^3 \\ \uparrow \\ (a, 0) \end{gathered} \end{aligned}$$
$$\begin{aligned} & \mathrm{at}=2 \mathrm{t}+\mathrm{t}^3 \\ & \mathrm{a}=2+\mathrm{t}^2 \\ & \mathrm{R}\left(2+\mathrm{t}^2, 0\right) \\ & \mathrm{PR}=4 \Rightarrow 4+4 \mathrm{t}^2=16 \end{aligned}$$
$$\begin{aligned} & \quad 4 \mathrm{t}^2=12 \Rightarrow \mathrm{t}^2=3 \\ & \mathrm{a}=5, \mathrm{R}(5,0) \end{aligned}$$
Focus $(1,0)$
$(1,0) \&(5,0)$ will be the end points of diameter
$\Rightarrow E q^{\mathrm{n}}$ of circle is
$$\begin{aligned} & (x-1)(x-5)+y^2=0 \\ & x^2+y^2-6 x+5=0 \end{aligned}$$
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