The radius of a circle which touches the y-ax | vedclass

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(C) Let the center of the circle be $(h, k)$.Since the circle touches the $y$-axis at $(0,3)$,the $y$-coordinate of the center is $k = 3$ and the radi

Vedclass · Accepted Answer

$5$

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(C) Let the center of the circle be $(h, k)$.Since the circle touches the $y$-axis at $(0,3)$,the $y$-coordinate of the center is $k = 3$ and the radius $r = |h|$.The equation of the circle is $(x - h)^2 + (y - 3)^2 = h^2$.This circle cuts an intercept of $8$ units on the $x$-axis. Setting $y = 0$ in the equation,we get $(x - h)^2 + (0 - 3)^2 = h^2$,which simplifies to $(x - h)^2 + 9 = h^2$,or $(x - h)^2 = h^2 - 9$.Thus,$x = h \pm \sqrt{h^2 - 9}$.The length of the intercept on the $x$-axis is the difference between these two $x$-values: $(h + \sqrt{h^2 - 9}) - (h - \sqrt{h^2 - 9}) = 2\sqrt{h^2 - 9}$.Given the intercept is $8$,we have $2\sqrt{h^2 - 9} = 8$,so $\sqrt{h^2 - 9} = 4$.Squaring both sides gives $h^2 - 9 = 16$,so $h^2 = 25$,which means $h = \pm 5$.Since the radius $r = |h|$,we have $r = 5$.

The radius of a circle which touches the $y$-axis at $(0,3)$ and cuts an intercept of $8$ units with the $x$-axis is

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