The number of common tangents that can be dra | vedclass

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(D) For the circle $x^2+y^2-6x=0$,the center $C_1 = (3, 0)$ and radius $r_1 = \sqrt{3^2+0^2-0} = 3$.For the circle $x^2+y^2+6x+2y+1=0$,the center $C_2

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$4$

Vedclass · Answer

(D) For the circle $x^2+y^2-6x=0$,the center $C_1 = (3, 0)$ and radius $r_1 = \sqrt{3^2+0^2-0} = 3$.For the circle $x^2+y^2+6x+2y+1=0$,the center $C_2 = (-3, -1)$ and radius $r_2 = \sqrt{(-3)^2+(-1)^2-1} = \sqrt{9+1-1} = 3$.The distance between the centers $d = C_1C_2 = \sqrt{(3 - (-3))^2 + (0 - (-1))^2} = \sqrt{6^2 + 1^2} = \sqrt{37}$.Since $r_1 + r_2 = 3 + 3 = 6$ and $\sqrt{36}  r_1 + r_2$.Because the distance between the centers is greater than the sum of the radii,the two circles lie outside each other without touching.Therefore,the number of common tangents that can be drawn is $4$.

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

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