The equation of the circle passing through (0 | vedclass

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

Question

(D) The circle passes through the origin $(0,0)$ and makes intercepts $a$ and $b$ on the $x$ and $y$ axes respectively.Therefore,the circle passes thr

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The circle passes through the origin $(0,0)$ and makes intercepts $a$ and $b$ on the $x$ and $y$ axes respectively.Therefore,the circle passes through the points $(a,0)$ and $(0,b)$.Since the angle between the coordinate axes is $90^{\circ}$,the line segment joining $(a,0)$ and $(0,b)$ is a diameter of the circle.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 $(x_1, y_1) = (a,0)$ and $(x_2, y_2) = (0,b)$:$(x-a)(x-0) + (y-0)(y-b) = 0$$x(x-a) + y(y-b) = 0$$x^2 - ax + y^2 - by = 0$$x^2 + y^2 - ax - by = 0$

The equation of the circle passing through $(0,0)$ and which makes intercepts $a$ and $b$ on the coordinate axes is

Similar Questions

The equation of the circle with centre on the $x$-axis,radius $5$ and passing through the point $(2, 3)$ is:

The equation of the circle which touches the lines $x = 0$,$y = 0$ and $3x + 4y = 4$ is

If the lines $3x - 4y - 7 = 0$ and $2x - 3y - 5 = 0$ are diameters of a circle with an area of $49\pi$ square units,then the equation of the circle is:

The equations of the circle which pass through the origin and make intercepts of lengths $4$ and $8$ on the $x$ and $y$-axes respectively are

$A$ circle is concentric with the circle $x^2 + y^2 - 6x + 12y + 15 = 0$ and has an area double that of the given circle. The equation of the circle is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module