Consider a circle C which touches the y-axis | vedclass

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(B) Let the center of the circle be $(h, k)$. Since the circle touches the $y$-axis at $(0, 6)$,the $y$-coordinate of the center is $k = 6$ and the ra

Vedclass · Accepted Answer

$9$

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(B) Let the center of the circle be $(h, k)$. Since the circle touches the $y$-axis at $(0, 6)$,the $y$-coordinate of the center is $k = 6$ and the radius $r = |h|$.Thus,the equation of the circle is $(x - h)^{2} + (y - 6)^{2} = h^{2}$.This circle cuts an intercept of $6 \sqrt{5}$ on the $x$-axis. Setting $y = 0$ in the equation,we get $(x - h)^{2} + (0 - 6)^{2} = h^{2}$,which simplifies to $(x - h)^{2} + 36 = h^{2}$,or $(x - h)^{2} = h^{2} - 36$.Taking the square root,$x - h = \pm \sqrt{h^{2} - 36}$,so $x = h \pm \sqrt{h^{2} - 36}$.The length of the intercept on the $x$-axis is the difference between these two $x$-values: $(h + \sqrt{h^{2} - 36}) - (h - \sqrt{h^{2} - 36}) = 2 \sqrt{h^{2} - 36}$.Given that the intercept is $6 \sqrt{5}$,we have $2 \sqrt{h^{2} - 36} = 6 \sqrt{5}$,so $\sqrt{h^{2} - 36} = 3 \sqrt{5}$.Squaring both sides,$h^{2} - 36 = 9 	imes 5 = 45$,so $h^{2} = 81$,which means $h = \pm 9$.Since the radius $r = |h|$,the radius of the circle is $r = 9$.

Consider a circle $C$ which touches the $y$-axis at $(0,6)$ and cuts off an intercept $6 \sqrt{5}$ on the $x$-axis. Then the radius of the circle $C$ is equal to:

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