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(C) Let the center of the circle be $O' = (h, k)$.Since the circle touches the $y$-axis at $(0, 4)$,the $y$-coordinate of the center is $k = 4$ and th

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$5$

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(C) Let the center of the circle be $O' = (h, k)$.Since the circle touches the $y$-axis at $(0, 4)$,the $y$-coordinate of the center is $k = 4$ and the radius $r = |h|$.Thus,the equation of the circle is $(x - h)^2 + (y - 4)^2 = h^2$.This circle cuts the $x$-axis $(y = 0)$ at points where $(x - h)^2 + (0 - 4)^2 = h^2$,which simplifies to $(x - h)^2 + 16 = h^2$,or $(x - h)^2 = h^2 - 16$.So,$x - h = \pm \sqrt{h^2 - 16}$,which gives $x = h \pm \sqrt{h^2 - 16}$.The length of the chord on the $x$-axis is the difference between these $x$-values: $(h + \sqrt{h^2 - 16}) - (h - \sqrt{h^2 - 16}) = 2\sqrt{h^2 - 16}$.Given the chord length is $6$,we have $2\sqrt{h^2 - 16} = 6$,so $\sqrt{h^2 - 16} = 3$.Squaring both sides,$h^2 - 16 = 9$,which gives $h^2 = 25$,so $h = 5$.Therefore,the radius $r = |h| = 5$.

$A$ circle touches the $y$-axis at the point $(0, 4)$ and cuts the $x$-axis in a chord of length $6$ units. The radius of the circle is

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