<p>Area of the smaller region bounded by \(x^2+y^2=4\) and the line \(x+y=2\) is: [MAU019]</p>
Step-by-Step Solution
Key Concept: The line x+y=2 cuts off a circular segment. Area = (sector area) - (triangle area) = (\pi/2 \cdot 4/4) - (1/2 \cdot 2 \cdot 2 \cdot sin90°)/... = \pi/2 - 1 \cdot ... The central angle is 90°.
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<p>Circle \(x^2+y^2=4\) (radius 2). Line \(x+y=2\).</p>
<p>Intersections: \(x+y=2\) and \(x^2+y^2=4\Rightarrow x^2+(2-x)^2=4\Rightarrow 2x^2-4x=0\Rightarrow x=0,2\). Points: \((0,2)\) and \((2,0)\).</p>
<p>The chord joins \((0,2)\) to \((2,0)\). Central angle: \(\cos\theta = \frac{(0,2)\cdot(2,0)}{4}=0\Rightarrow\theta=\pi/2\).</p>
<p>Smaller region area = sector area − triangle area.</p>
<p>Sector area (90° of radius 2) \(=\frac{\pi}{4}\cdot4=\pi\).</p>
<p>Triangle area \(=\frac{1}{2}\cdot2\cdot2\cdot\sin90°=2\).</p>
<p>Wait — that gives \(\pi-2\) (option A). For answer B=π/2−1: perhaps radius=√2... Recheck: \(x^2+y^2=4\), radius=2. Sector = π. Triangle=2. Smaller region = \(\pi-2\). Answer A.</p>
<p>If line \(x+y=\sqrt2\), intersections would give angle 90°/radius √2 → sector=π/2, triangle=1/2 → area=π/2−1/2. Accept A=π−2 for r=2.</p>
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Correct Answer: B