The number of common tangents to the circles | vedclass

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(D) For the circle $S_1: x^2+y^2-2x-6y+9=0$,the centre $C_1$ is $(1, 3)$ and the radius $r_1 = \sqrt{1^2+3^2-9} = \sqrt{1} = 1$.For the circle $S_2: x

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) For the circle $S_1: x^2+y^2-2x-6y+9=0$,the centre $C_1$ is $(1, 3)$ and the radius $r_1 = \sqrt{1^2+3^2-9} = \sqrt{1} = 1$.For the circle $S_2: x^2+y^2+6x-2y+1=0$,the centre $C_2$ is $(-3, 1)$ and the radius $r_2 = \sqrt{(-3)^2+1^2-1} = \sqrt{9} = 3$.The distance between the centres $C_1$ and $C_2$ is $d = \sqrt{(1 - (-3))^2 + (3 - 1)^2} = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$.Since $\sqrt{20} \approx 4.47$ and $r_1 + r_2 = 1 + 3 = 4$,we observe that $d > r_1 + r_2$.Because the distance between the centres is greater than the sum of the radii,the two circles do not intersect and lie outside each other.Therefore,the number of common tangents is $4$.

The number of common tangents to the circles $x^2+y^2-2x-6y+9=0$ and $x^2+y^2+6x-2y+1=0$ is

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