JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Circle

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 AB and passing through the mid point of the chord AB intersects is the circle at C and D , then the area of the quadrilateral ADBC is equal to

(A) $37$
(B) $214$ ✓ Correct
(C) $57$
(D) $14$

Answer: (B)

Solution

The line

x + y = 1

meets the circle

x^{2} + y^{2} = 4

A, B

; the distance between them follows from

A B^{2} = 4 (R^{2} - d^{2}), R = 2, d = \frac{∣0 + 0 - 1∣}{2} = \frac{1}{2},

giving

A B^{2} = 4 (4 - \frac{1}{2}) = 14

.

The line perpendicular to

A B

through the midpoint of

A B

is the perpendicular bisector of chord

A B

, which necessarily passes through the centre

O

; hence

C D

is a diameter,

C D = 2 R = 4

.

Since the diagonals

A B

and

C D

of quadrilateral

A D B C

are perpendicular, its area is

\frac{1}{2} \cdot A B \cdot C D = \frac{1}{2} \cdot 14 \cdot 4 = 214 .

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Single correct option

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JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Circle

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