Let the line x + y = 1 meet the circle at the points A and B. If the line perpendicular to AB and passing through the mid point of the chord AB intersects the circle at C and D the area of the quadrilateral ADBC is equal to: [2025]

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Solution

Q.
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 the circle at C and D, then the area of the quadrilateral ADBC is equal to: [2025]

1 $3 \sqrt{7}$

2 $\sqrt{14}$

3 $2 \sqrt{14}$

4 $5 \sqrt{7}$

Ans.
(3)

For points of intersection A and B.

Solving x + y = 1 and $x^{2} + y^{2} = 4$ , we get

$A (\frac{1 - \sqrt{7}}{2}, \frac{1 + \sqrt{7}}{2}), B (\frac{1 + \sqrt{7}}{2}, \frac{1 - \sqrt{7}}{2})$

Slope of AB = –1

Slope of $⊥^{r}$ bisector of AB = 1

For points of intersection C and D

Solving, x = y and $x^{2} + y^{2} = 4$ , we get

$C (\sqrt{2}, \sqrt{2})$ and $D (- \sqrt{2}, - \sqrt{2})$

$∴$ Area of quadrilateral ADBC = 2 $\times$ Area of $△$ BCD

$= 2 \times \frac{1}{2}$ $| \begin{matrix} \sqrt{2} & \sqrt{2} & 1 \\ \frac{1 + \sqrt{7}}{2} & \frac{1 - \sqrt{7}}{2} & 1 \\ - \sqrt{2} & - \sqrt{2} & 1 \end{matrix} |$ $= 2 \sqrt{14}$ sq. units.

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