Consider the curve \(x^2+y^2=1\) and the line \(x+y=1\).
Which one of the following shaded portions represents the area bounded by the curve \(x^2+y^2 \le 1\) and \(x+y \gt 1\)?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) →
\(x^2+y^2 \le 1\) is the region inside the circle with centre (0, 0) and radius 1.
The line \(x+y=1\) meets the circle where
\( x^2+(1-x)^2 = 1 \)
\( 2x^2-2x = 0 \), so \( x = 0 \) or \( x = 1 \)
that is, at (0, 1) and (1, 0).
To find the side of the line for \(x+y \gt 1\), test the origin: \(0+0 = 0\), which is not greater than 1. So \(x+y \gt 1\) is the side of the line away from the origin.
The region is therefore the small part of the circle in the first quadrant cut off by the chord from (1, 0) to (0, 1), on the side away from the origin. Option (2) shades exactly this part.
Why the other options are incorrect
Option (1) — Incorrect. It shades the triangle between the axes and the chord, on the origin side of the line, where \(x+y \lt 1\).
Option (3) — Incorrect. It shades a part of the circle in the second quadrant. There \(x \le 0\) and \(y \le 1\), so \(x+y \le 1\), and the condition \(x+y \gt 1\) fails.
Option (4) — Incorrect. It shades a triangle in the second quadrant, where again \(x \le 0\) and \(y \le 1\), so \(x+y \gt 1\) fails.