Number of common tangents to the circles x^2+ | vedclass

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(C) For the circle $x^2+y^2-6x-14y+48=0$,the center $C_1$ is $(3, 7)$ and the radius $r_1 = \sqrt{3^2+7^2-48} = \sqrt{9+49-48} = \sqrt{10}$.For the ci

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$4$

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(C) For the circle $x^2+y^2-6x-14y+48=0$,the center $C_1$ is $(3, 7)$ and the radius $r_1 = \sqrt{3^2+7^2-48} = \sqrt{9+49-48} = \sqrt{10}$.For the circle $x^2+y^2-6x=0$,the center $C_2$ is $(3, 0)$ and the radius $r_2 = \sqrt{3^2+0^2-0} = 3$.The distance between the centers $C_1$ and $C_2$ is $d = \sqrt{(3-3)^2+(7-0)^2} = 7$.Since $r_1 + r_2 = \sqrt{10} + 3 \approx 3.16 + 3 = 6.16$,we observe that $d > r_1 + r_2$ because $7 > 6.16$.Since the distance between the centers is greater than the sum of the radii,the circles are disjoint and lie outside each other.Therefore,the number of common tangents is $4$.

Number of common tangents to the circles $x^2+y^2-6x-14y+48=0$ and $x^2+y^2-6x=0$ are

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