Let S be a subset of the plane defined by S = | vedclass

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

Question

(B) The set $S = \{(x, y) : |x| + 2|y| = 1\}$ represents a rhombus in the Cartesian plane with vertices at $(1, 0), (-1, 0), (0, 1/2),$ and $(0, -1/2)

Vedclass · Accepted Answer

$\frac{1}{\sqrt{5}}$

Vedclass · Answer

(B) The set $S = \{(x, y) : |x| + 2|y| = 1\}$ represents a rhombus in the Cartesian plane with vertices at $(1, 0), (-1, 0), (0, 1/2),$ and $(0, -1/2)$.The smallest circle with centre at the origin $(0, 0)$ that has a non-empty intersection with $S$ is the circle inscribed in this rhombus.The radius $r$ of this inscribed circle is the perpendicular distance from the origin $(0, 0)$ to any of the sides of the rhombus.Consider the side in the first quadrant,which is given by the line $x + 2y = 1$ or $x + 2y - 1 = 0$.The perpendicular distance $r$ from the origin $(0, 0)$ to the line $Ax + By + C = 0$ is given by $r = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.Substituting $A = 1, B = 2, C = -1$ and $(x_0, y_0) = (0, 0)$:$r = \frac{|1(0) + 2(0) - 1|}{\sqrt{1^2 + 2^2}} = \frac{|-1|}{\sqrt{1 + 4}} = \frac{1}{\sqrt{5}}$.

Let $S$ be a subset of the plane defined by $S = \{(x, y) : |x| + 2|y| = 1\}$. Then,the radius of the smallest circle with centre at the origin and having non-empty intersection with $S$ is

Similar Questions

The area of the curve $x^2 + y^2 = 2ax$ is

Which of the following lines is a diameter of the circle $x^2 + y^2 - 6x - 8y - 9 = 0$?

Find the equation of the circle which touches the $X$-axis at a distance of $+3$ from the origin and cuts an intercept of $8$ on the positive $Y$-axis.

Three circles,each of radius $1$,touch one another externally and lie between two parallel lines. The minimum possible distance between the lines is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module