Q53 / 150 · MHT CET 2026 (16 April Shift 2) · Mathematics
Line $l : x + y = 4$ intersects the circle $x^2 + y^2 - 2x - 2y = 2$ at points $A$ and $B$. If C is the center of the circle, then the area of $\triangle ABC$ is...
- A$\sqrt{2}$
- Correct answerB$2$
- C$2\sqrt{2}$
- D$4$
Solution
Equation of the circle is $x^2 + y^2 - 2x - 2y = 2$, which can be written as $(x-1)^2 + (y-1)^2 = 4$.
Center of the circle is $C(1, 1)$ and radius is $R = 2$.
The perpendicular distance $d$ from the center $C(1, 1)$ to the line $x + y - 4 = 0$ is given by $d = \dfrac{|1 + 1 - 4|}{\sqrt{1^2 + 1^2}} = \sqrt{2}$.
The length of the chord $AB$ is $2\sqrt{R^2 - d^2} = 2\sqrt{2^2 - (\sqrt{2})^2} = 2\sqrt{2}$.
The area of $\triangle ABC$ is $\dfrac{1}{2} \times AB \times d = \dfrac{1}{2} \times 2\sqrt{2} \times \sqrt{2} = 2$.
Answer: $2$
Center of the circle is $C(1, 1)$ and radius is $R = 2$.
The perpendicular distance $d$ from the center $C(1, 1)$ to the line $x + y - 4 = 0$ is given by $d = \dfrac{|1 + 1 - 4|}{\sqrt{1^2 + 1^2}} = \sqrt{2}$.
The length of the chord $AB$ is $2\sqrt{R^2 - d^2} = 2\sqrt{2^2 - (\sqrt{2})^2} = 2\sqrt{2}$.
The area of $\triangle ABC$ is $\dfrac{1}{2} \times AB \times d = \dfrac{1}{2} \times 2\sqrt{2} \times \sqrt{2} = 2$.
Answer: $2$
Hence, the correct answer is option: 'B'.