Four distinct points (2k, 3k), (1, 0), (0, 1) | vedclass

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

Question

(C) The points $(0, 0), (1, 0),$ and $(0, 1)$ form a right-angled triangle with the right angle at the origin $(0, 0)$.Since these three points lie on

Vedclass · Accepted Answer

$\frac{5}{13}$

Vedclass · Answer

(C) The points $(0, 0), (1, 0),$ and $(0, 1)$ form a right-angled triangle with the right angle at the origin $(0, 0)$.Since these three points lie on a circle,the segment connecting the points $(1, 0)$ and $(0, 1)$ 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 $(1, 0)$ and $(0, 1)$ as diameter endpoints:$(x - 1)(x - 0) + (y - 0)(y - 1) = 0$$x(x - 1) + y(y - 1) = 0$$x^2 - x + y^2 - y = 0$$x^2 + y^2 - x - y = 0$For the point $(2k, 3k)$ to lie on this circle,it must satisfy the equation:$(2k)^2 + (3k)^2 - (2k) - (3k) = 0$$4k^2 + 9k^2 - 5k = 0$$13k^2 - 5k = 0$$k(13k - 5) = 0$This gives $k = 0$ or $k = \frac{5}{13}$.Since the points must be distinct,$k$ cannot be $0$ (as that would make the point $(0, 0)$,which is already given). Therefore,$k = \frac{5}{13}$.

Four distinct points $(2k, 3k), (1, 0), (0, 1),$ and $(0, 0)$ lie on a circle for $k$ equal to:

Similar Questions

The parametric equations of the circle $x^2+y^2-ax-by=0$ are

The equation of a circle whose center lies on $x + 2y = 0$ and which touches the lines $3x - 4y + 8 = 0$ and $3x - 4y - 28 = 0$ is

The equation of the circumcircle of the quadrilateral formed by the lines $x = a, x = 2a, y = -a, y = a$ is:

The equation of the circle concentric with the circle $x^2+y^2-6x+12y+15=0$ and having double its area is:

The centre and radius of the circle $(a + 1)x^2 + 3y^2 - 6x + 9y + a + 4 = 0$ are respectively ...

Vedclass Test Series

Exam Paper Generator

Online Exam Module