The line l : x + y = 4 intersects the circle | vedclass

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(B) $1$. The equation of the circle is $x^2 + y^2 - 2x - 2y - 2 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$, we get $g = -1, f = -1, c = -2$.

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$2$

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(B) $1$. The equation of the circle is $x^2 + y^2 - 2x - 2y - 2 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$, we get $g = -1, f = -1, c = -2$.$2$. The center $C$ is $(-g, -f) = (1, 1)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{1 + 1 + 2} = \sqrt{4} = 2$.$3$. The perpendicular distance $d$ from $C(1, 1)$ to the line $x + y - 4 = 0$ is $d = \frac{|1 + 1 - 4|}{\sqrt{1^2 + 1^2}} = \frac{|-2|}{\sqrt{2}} = \sqrt{2}$.$4$. In $	riangle ABC$, $AC = BC = r = 2$. The height of the triangle from $C$ to chord $AB$ is $d = \sqrt{2}$.$5$. The length of the chord $AB = 2\sqrt{r^2 - d^2} = 2\sqrt{4 - 2} = 2\sqrt{2}$.$6$. Area of $	riangle ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes AB 	imes d = \frac{1}{2} 	imes 2\sqrt{2} 	imes \sqrt{2} = 2$.

The 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...

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