The area of the region {(x,y) : x^2 + y^2 leq | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The region is defined by two inequalities: $1$. $x^2 + y^2 \leqslant 1$ (the interior of a circle centered at the origin with radius $1$). $2$. $x

Vedclass · Accepted Answer

$\frac{\pi}{4} - \frac{1}{2}$

Vedclass · Answer

(A) The region is defined by two inequalities: $1$. $x^2 + y^2 \leqslant 1$ (the interior of a circle centered at the origin with radius $1$). $2$. $x + y \geqslant 1$ (the region above or on the line $x + y = 1$). The intersection of these two regions is a circular segment. The line $x + y = 1$ passes through $(1, 0)$ and $(0, 1)$. The distance from the origin $(0, 0)$ to the line $x + y - 1 = 0$ is $d = \frac{|0 + 0 - 1|}{\sqrt{1^2 + 1^2}} = \frac{1}{\sqrt{2}}$. In a circle of radius $r = 1$,the area of a circular segment cut by a chord at distance $d$ from the center is given by $A = r^2 \cos^{-1}(\frac{d}{r}) - d\sqrt{r^2 - d^2}$. Substituting $r = 1$ and $d = \frac{1}{\sqrt{2}}$: $A = (1)^2 \cos^{-1}(\frac{1/\sqrt{2}}{1}) - \frac{1}{\sqrt{2}} \sqrt{1 - (\frac{1}{\sqrt{2}})^2}$ $A = \cos^{-1}(\frac{1}{\sqrt{2}}) - \frac{1}{\sqrt{2}} \sqrt{1 - \frac{1}{2}}$ $A = \frac{\pi}{4} - \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{\pi}{4} - \frac{1}{2}$.

The area of the region $\{(x,y) : x^2 + y^2 \leqslant 1 \leqslant x + y\}$ is

Similar Questions

Let $C: x^2+y^2=4$ and $C^{\prime}: x^2+y^2-4 \lambda x+9=0$ be two circles. If the set of all values of $\lambda$ such that the circles $C$ and $C^{\prime}$ intersect at two distinct points is $\mathbb{R}-[a, b]$,then the point $(8a+12, 16b-20)$ lies on the curve:

Let $AB$ be a line segment of length $2$. Construct a semicircle $S$ with $AB$ as diameter. Let $C$ be the mid-point of the arc $AB$. Construct another semicircle $T$ external to the $\triangle ABC$ with chord $AC$ as diameter. The area of the region inside the semicircle $T$ but outside $S$ is

Find the circumcenter of the triangle formed by the points $(a \cos \alpha, a \sin \alpha)$,$(a \cos \beta, a \sin \beta)$,and $(a \cos \gamma, a \sin \gamma)$.

Let $ABC$ be a triangle with $AB=1$,$AC=3$,and $\angle BAC=\frac{\pi}{2}$. If a circle of radius $r>0$ touches the sides $AB$,$AC$ and also touches the circumcircle of triangle $ABC$ internally,then the value of $r$ is:

If two circles $(x - 1)^2 + (y - 3)^2 = r^2$ and $x^2 + y^2 - 8x + 2y + 8 = 0$ intersect in two distinct points,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module