If x^2+y^2+6x+2ky+25=0 touches the Y-axis,the | vedclass

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

Question

(C) The given equation of the circle is $x^2+y^2+6x+2ky+25=0$. Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $g=3$,$f=k$,and $c=25

Vedclass · Accepted Answer

$\pm 5$

Vedclass · Answer

(C) The given equation of the circle is $x^2+y^2+6x+2ky+25=0$. Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $g=3$,$f=k$,and $c=25$. The center of the circle is $(-g, -f) = (-3, -k)$ and the radius $r$ is given by $\sqrt{g^2+f^2-c} = \sqrt{3^2+k^2-25} = \sqrt{k^2-16}$. Since the circle touches the $Y$-axis,the radius must be equal to the absolute value of the $x$-coordinate of the center,i.e.,$r = |-g| = |-3| = 3$. Equating the two expressions for the radius: $\sqrt{k^2-16} = 3$. Squaring both sides: $k^2-16 = 9$. $k^2 = 25$,which gives $k = \pm 5$.

If $x^2+y^2+6x+2ky+25=0$ touches the $Y$-axis,then $k=$

Similar Questions

The centre of the circle passing through the point $(0,1)$ and touching the curve $y=x^2$ at $(2,4)$ is

Find the equation of the circle with centre $(0,2)$ and radius $2$.

The equation of the circle with centre on the $x$-axis,radius $4$ and passing through the origin,is

Find the equation of the circle passing through the point $(3, 6)$ and having its center at $(2, -1)$.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module