The equation of the circle passing through (1 | vedclass

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

Question

(C) For a circle passing through two points to have the smallest possible radius,the line segment joining these two points must be the diameter of the

Vedclass · Accepted Answer

$x^2 + y^2 - x - y = 0$

Vedclass · Answer

(C) For a circle passing through two points to have the smallest possible radius,the line segment joining these two points must be the diameter of the circle.Given points are $A(1, 0)$ and $B(0, 1)$.The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is given by $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.Substituting the given points:$(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$

The equation of the circle passing through $(1,0)$ and $(0,1)$ and having the smallest possible radius is:

Similar Questions

The area of the circle whose centre is at $(1, 2)$ and which passes through the point $(4, 6)$ is

The least distance of the point $A(10, 7)$ from the circle $x^2 + y^2 - 4x - 2y - 20 = 0$ is the length of segment $AM$. If $MM'$ is the diameter of the circle,then the lengths of $AM$ and $AM'$ are respectively . . . . . . , . . . . . . units.

The length of the shortest path that begins at the point $(2, 5)$,touches the $x-$axis,and then ends at a point on the circle $x^2 + y^2 + 12x - 20y + 120 = 0$.

The number of circles having radius $5$ and passing through the points $(-2, 0)$ and $(4, 0)$ is

Find the number of common tangents that can be drawn to the circles $x^2 + y^2 - 4x - 6y - 3 = 0$ and $x^2 + y^2 + 2x + 2y + 1 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module