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

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(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

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$\pm 5$

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(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 $k=$

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