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(D) The given curves are a hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ and a circle $x^2+y^2=16$. For the hyperbola,$a^2 = 16$ and $b^2 = 9$. The equat

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$2$

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(D) The given curves are a hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ and a circle $x^2+y^2=16$. For the hyperbola,$a^2 = 16$ and $b^2 = 9$. The equation of any tangent to the hyperbola is $y = mx \pm \sqrt{a^2m^2 - b^2}$,which is $y = mx \pm \sqrt{16m^2 - 9}$. For this line to be a tangent to the circle $x^2+y^2=16$ (which has center $(0,0)$ and radius $r=4$),the perpendicular distance from the center to the line must equal the radius. Thus,$\frac{|\pm \sqrt{16m^2 - 9}|}{\sqrt{m^2+1}} = 4$. Squaring both sides,we get $\frac{16m^2 - 9}{m^2+1} = 16$. $16m^2 - 9 = 16m^2 + 16$. $-9 = 16$,which is impossible. This implies there are no real values of $m$ for which the tangent to the hyperbola is also tangent to the circle. However,we must check for vertical tangents. The hyperbola has vertical tangents at $x = \pm 4$. The circle $x^2+y^2=16$ also has vertical tangents at $x = \pm 4$. Thus,the lines $x=4$ and $x=-4$ are common tangents to both curves. Therefore,there are $2$ common tangents.

The number of common tangents that can be drawn to the curves $\frac{x^2}{16}-\frac{y^2}{9}=1$ and $x^2+y^2=16$ is

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