If the circles x^2+y^2=9 and x^2+y^2-8x-6y+n^ | vedclass

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(C) The first circle is $C_1: x^2+y^2=3^2$,with center $O_1(0,0)$ and radius $r_1=3$. The second circle is $C_2: x^2+y^2-8x-6y+n^2=0$,which can be wri

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$9$

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(C) The first circle is $C_1: x^2+y^2=3^2$,with center $O_1(0,0)$ and radius $r_1=3$. The second circle is $C_2: x^2+y^2-8x-6y+n^2=0$,which can be written as $(x-4)^2+(y-3)^2 = 25-n^2$. Thus,the center is $O_2(4,3)$ and the radius is $r_2 = \sqrt{25-n^2}$. For the circles to have exactly two common tangents,they must intersect at two distinct points,which occurs when $|r_1-r_2| Here,$d = \sqrt{(4-0)^2+(3-0)^2} = \sqrt{16+9} = 5$. The condition $|3-\sqrt{25-n^2}| From $5  2$,so $25-n^2 > 4$,which means $n^2 From $|3-\sqrt{25-n^2}| This implies $\sqrt{25-n^2}  -2$ (always true). Also,for the radius to be real,$25-n^2 > 0$,so $n^2 Combining $n^2 Since $n \in \mathbb{Z}$,$n^2 \in \{0, 1, 4, 9, 16\}$. Possible values for $n$ are $0, \pm 1, \pm 2, \pm 3, \pm 4$. Counting these,we have $1 + 2 + 2 + 2 + 2 = 9$ values.

If the circles $x^2+y^2=9$ and $x^2+y^2-8x-6y+n^2=0$,where $n \in \mathbb{Z}$,have exactly two common tangents,then the number of values for $n$ is

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