The equation of the circle passing through th | vedclass

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

Question

(A) circle passing through the origin $(0,0)$ and making intercepts of length $a = 6$ on the $x$-axis and $b = 4$ on the $y$-axis has its center at $(

Vedclass · Accepted Answer

$x^2 + y^2 - 6x - 4y = 0$

Vedclass · Answer

(A) circle passing through the origin $(0,0)$ and making intercepts of length $a = 6$ on the $x$-axis and $b = 4$ on the $y$-axis has its center at $(\frac{a}{2}, \frac{b}{2})$.Since the center lies in the first quadrant,the coordinates are $(\frac{6}{2}, \frac{4}{2}) = (3, 2)$.The general equation of a circle passing through the origin with center $(h, k)$ is $x^2 + y^2 - 2hx - 2ky = 0$.Substituting $h = 3$ and $k = 2$,we get $x^2 + y^2 - 2(3)x - 2(2)y = 0$.Thus,the required equation is $x^2 + y^2 - 6x - 4y = 0$.

The equation of the circle passing through the origin,whose center lies in the first quadrant and which makes intercepts of length $6$ and $4$ on the $x$-axis and $y$-axis respectively,is:

Similar Questions

If the lines $x + y = 6$ and $x + 2y = 4$ are diameters of a circle whose diameter is $20$,then the equation of the circle is:

If $P_1, P_2, P_3$ are the perimeters of the three circles $x^2+y^2+8x-6y=0$,$4x^2+4y^2-4x-12y-186=0$,and $x^2+y^2-6x+6y-9=0$ respectively,then

The equation of the circle whose diameter lines are $2x + 3y = 3$ and $16x - y = 4$ and which passes through the point $(4, 6)$ is

The equation of the circle whose centre is $(1, -3)$ and which touches the line $2x - y - 4 = 0$ is

If three distinct points $A, B, C$ are given in the $2$-dimensional coordinate plane such that the ratio of the distance from $(1, 0)$ to the distance from $(-1, 0)$ is equal to $\frac{1}{2}$,then the circumcentre of the triangle $ABC$ is at the point

Vedclass Test Series

Exam Paper Generator

Online Exam Module