JEE Main 2025 Jan 29 Shift 1, Mathematics Q1: Circle: Chord, Tangent And Normal

JEE Main2025Jan 29, Shift 1Mathematics

Let the line $x + y = 1$ meet the circle $x^{2} + y^{2} = 4$ at the points $A$ and $B$ . If the line perpendicular to $A B$ and passing through the mid point of the chord $A B$ intersects the circle at $C$ and $D$ , then the area of the quadrilateral $A D B C$ is equal to

A
$37$
B
$214$
C
$57$
D
$14$

Solution

The chord $A B$ lies on $x + y = 1$ and the circle is centred at the origin. The perpendicular to $A B$ through the midpoint of $A B$ passes through the centre, so $C D$ is a diameter along the line $y = x$ .

Chord $A B$ : substituting $y = 1 - x$ in $x^{2} + y^{2} = 4$ gives $2 x^{2} - 2 x - 3 = 0$ , hence $x = \frac{1 \pm 7}{2}$ . So $∣ A B ∣^{2} = 2 (x_{1} - x_{2})^{2} = 2 \cdot 7 = 14$ , giving $∣ A B ∣ = 14$ .

Diameter $C D$ : intersection of $y = x$ with the circle gives $C (2, 2)$ and $D (- 2, - 2)$ , so $∣ C D ∣ = 4$ .

The diagonals $A B$ and $C D$ are perpendicular (slopes $- 1$ and $+ 1$ ), so the quadrilateral $A D B C$ has area $\frac{1}{2} ∣ A B ∣∣ C D ∣ = \frac{1}{2} \cdot 14 \cdot 4 = 214 .$

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JEE Main 2025 Jan 29 Shift 1, Mathematics Q1: Circle: Chord, Tangent And Normal

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