The circle x^2 + y^2 + 6y = 0 touches which o | vedclass

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

Question

(B) The given equation of the circle is $x^2 + y^2 + 6y = 0$.Completing the square for the $y$-terms:$x^2 + (y^2 + 6y + 9) = 9$$x^2 + (y + 3)^2 = 3^2$

Vedclass · Accepted Answer

$x$-axis at the origin

Vedclass · Answer

(B) The given equation of the circle is $x^2 + y^2 + 6y = 0$.Completing the square for the $y$-terms:$x^2 + (y^2 + 6y + 9) = 9$$x^2 + (y + 3)^2 = 3^2$Comparing this with the standard form $(x - h)^2 + (y - k)^2 = r^2$,the center is $(0, -3)$ and the radius $r = 3$.Since the distance from the center $(0, -3)$ to the $x$-axis is $|-3| = 3$,which is equal to the radius,the circle touches the $x$-axis.Since the $x$-coordinate of the center is $0$,the circle touches the $x$-axis at the origin $(0, 0)$.

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

Similar Questions

The circle passing through the point $(-1, 0)$ and touching the $y$-axis at $(0, 2)$ also passes through the point

The equation of the circle passing through the point $(2, 1)$ and touching the $y$-axis at the origin is:

The equation of the circle concentric with the circle $x^2 + y^2 + 8x + 10y - 7 = 0$ and passing through the centre of the circle $x^2 + y^2 - 4x - 6y = 0$ is

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

The coordinates of any point on the circle $x^2 + y^2 = 4$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module