JEE Main 2025 Jan 23 Shift 2, Mathematics Q13: Circle: Chord, Tangent And Normal

JEE Main2025Jan 23, Shift 2Mathematics

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 :

A
$x^{2} + y^{2} - 6 x + 5 = 0$
B
$x^{2} + y^{2} - 4 x + 3 = 0$
C
$x^{2} + y^{2} - 10 x + 9 = 0$
D
$x^{2} + y^{2} - 8 x + 7 = 0$

Solution

For $y^{2} = 4 x$ , parametrise points as $P (t^{2}, 2 t)$ . The normal at $P$ has the form $y + t x = 2 t + t^{3}$ .

The point $(a, 0)$ lies on this normal: $0 + a t = 2 t + t^{3} \Rightarrow a = 2 + t^{2}$ (for $t \neq = 0$ ).

Shortest distance: $dist ((a, 0), P) = (a - t^{2})^{2} + (2 t)^{2} = 4 + 4 t^{2} = 2 1 + t^{2}$ .

Setting this equal to $4$ : $1 + t^{2} = 4 \Rightarrow t^{2} = 3 \Rightarrow a = 5$ .

The focus of $y^{2} = 4 x$ is $(1, 0)$ . We need the circle through $(5, 0)$ and $(1, 0)$ with centre on the $x$ -axis. The two given points are then endpoints of a diameter (centre is the midpoint $(3, 0)$ , radius $2$ ).

Diameter form: $(x - 1) (x - 5) + y^{2} = 0 \Rightarrow x^{2} + y^{2} - 6 x + 5 = 0$ .

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JEE Main 2025 Jan 23 Shift 2, Mathematics Q13: Circle: Chord, Tangent And Normal

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