The total number of common tangents of x^{2}+ | vedclass

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(C) The given equations of the circles are:$C_1: x^2 + y^2 - 6x - 8y + 9 = 0$$C_2: x^2 + y^2 = 1$For $C_1$,the center is $(3, 4)$ and the radius $R_1

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$3$

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(C) The given equations of the circles are:$C_1: x^2 + y^2 - 6x - 8y + 9 = 0$$C_2: x^2 + y^2 = 1$For $C_1$,the center is $(3, 4)$ and the radius $R_1 = \sqrt{3^2 + 4^2 - 9} = \sqrt{9 + 16 - 9} = \sqrt{16} = 4$.For $C_2$,the center is $(0, 0)$ and the radius $R_2 = 1$.The distance between the centers $C_1(3, 4)$ and $C_2(0, 0)$ is $d = \sqrt{(3-0)^2 + (4-0)^2} = \sqrt{9 + 16} = 5$.We observe that $R_1 + R_2 = 4 + 1 = 5$.Since the distance between the centers $d = R_1 + R_2$,the two circles touch each other externally.When two circles touch each other externally,they have $3$ common tangents (two direct common tangents and one transverse common tangent).

The total number of common tangents of $x^{2}+y^{2}-6x-8y+9=0$ and $x^{2}+y^{2}=1$ is

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