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(B) Let the radius of the required circle be $r$. Since it touches the $y-$axis at $(0,2)$,its center is $(r, 2)$.The circle $x^2 + y^2 = 16$ has cent

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$\frac{3}{2}$

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(B) Let the radius of the required circle be $r$. Since it touches the $y-$axis at $(0,2)$,its center is $(r, 2)$.The circle $x^2 + y^2 = 16$ has center $O(0,0)$ and radius $R = 4$.Since the circles touch internally,the distance between their centers must be equal to the difference of their radii.The distance between centers $(r, 2)$ and $(0, 0)$ is $\sqrt{r^2 + 2^2} = \sqrt{r^2 + 4}$.For internal contact,$\sqrt{r^2 + 4} = R - r = 4 - r$.Squaring both sides: $r^2 + 4 = (4 - r)^2$.$r^2 + 4 = 16 - 8r + r^2$.$8r = 12$.$r = \frac{12}{8} = \frac{3}{2}$.

Find the radius of a circle that touches the $y-$axis at point $P(0,2)$ and touches the circle $x^2 + y^2 = 16$ internally.

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