If four distinct points (2k, 3k),(2,0),(0,3), | vedclass

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

Question

(C) The points $(2,0)$,$(0,3)$,and $(0,0)$ form a right-angled triangle at the origin $(0,0)$ because the angle between the $x$-axis and $y$-axis is $

Vedclass · Accepted Answer

$k = 1$

Vedclass · Answer

(C) The points $(2,0)$,$(0,3)$,and $(0,0)$ form a right-angled triangle at the origin $(0,0)$ because the angle between the $x$-axis and $y$-axis is $90^{\circ}$.Since the angle subtended by the line segment joining $(2,0)$ and $(0,3)$ at the point $(0,0)$ is $90^{\circ}$,this line segment must be the diameter of the circle.The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.Substituting the points $(2,0)$ and $(0,3)$,we get:$(x-2)(x-0) + (y-0)(y-3) = 0$$x^2 - 2x + y^2 - 3y = 0$$x^2 + y^2 - 2x - 3y = 0$Since the point $(2k, 3k)$ lies on this circle,it must satisfy the equation:$(2k)^2 + (3k)^2 - 2(2k) - 3(3k) = 0$$4k^2 + 9k^2 - 4k - 9k = 0$$13k^2 - 13k = 0$$13k(k - 1) = 0$This gives $k = 0$ or $k = 1$.If $k = 0$,the point $(2k, 3k)$ becomes $(0,0)$,which is not a distinct point from the given point $(0,0)$.Therefore,for the four points to be distinct,we must have $k = 1$.

If four distinct points $(2k, 3k)$,$(2,0)$,$(0,3)$,and $(0,0)$ lie on a circle,then:

Similar Questions

The equation of the circle whose centre lies on the lines $3x - y - 4 = 0$ and $x + 3y + 2 = 0$ and has an area of $154$ square units is:

The equation of a circle that passes through the origin and cuts off intercepts $-2$ and $3$ on the $X$-axis and $Y$-axis respectively is

The points of contact of the circle $x^2 + y^2 + 2x + 2y + 1 = 0$ with the coordinate axes are:

The circle $x^2 + y^2 + 6y = 0$ touches which of the following?

Find the equation of the circle with centre $(0,2)$ and radius $2$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module