The number of common tangents that can be dra | vedclass

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(A) For the circle $x^2+y^2=1$:Centre $C_1 = (0,0)$,Radius $R_1 = 1$.For the circle $x^2+y^2-2x-6y+6=0$:Centre $C_2 = (1,3)$,Radius $R_2 = \sqrt{1^2+3

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) For the circle $x^2+y^2=1$:Centre $C_1 = (0,0)$,Radius $R_1 = 1$.For the circle $x^2+y^2-2x-6y+6=0$:Centre $C_2 = (1,3)$,Radius $R_2 = \sqrt{1^2+3^2-6} = \sqrt{4} = 2$.The distance between the centres $C_1$ and $C_2$ is $d = \sqrt{(1-0)^2 + (3-0)^2} = \sqrt{1+9} = \sqrt{10}$.Since $\sqrt{10} \approx 3.16$ and $R_1+R_2 = 1+2 = 3$,we have $d > R_1+R_2$.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,there are $4$ common tangents (two direct and two transverse).

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

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