The number of common tangents to the circles | vedclass

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Question

(A) For the first circle $x^2 + y^2 - 8x - 2y + 1 = 0$,the center $C_1 = (4, 1)$ and radius $r_1 = \sqrt{4^2 + 1^2 - 1} = \sqrt{16 + 1 - 1} = \sqrt{16

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) For the first circle $x^2 + y^2 - 8x - 2y + 1 = 0$,the center $C_1 = (4, 1)$ and radius $r_1 = \sqrt{4^2 + 1^2 - 1} = \sqrt{16 + 1 - 1} = \sqrt{16} = 4$.For the second circle $x^2 + y^2 + 6x + y = 0$,the center $C_2 = (-3, -0.5)$ and radius $r_2 = \sqrt{(-3)^2 + (-0.5)^2 - 0} = \sqrt{9 + 0.25} = \sqrt{9.25} \approx 3.04$.The distance between the centers $d = \sqrt{(4 - (-3))^2 + (1 - (-0.5))^2} = \sqrt{7^2 + 1.5^2} = \sqrt{49 + 2.25} = \sqrt{51.25} \approx 7.16$.Since $d = \sqrt{51.25} > r_1 + r_2 = 4 + 3.04 = 7.04$,the circles lie outside each other.Therefore,the number of common tangents is $4$.

The number of common tangents to the circles $x^2 + y^2 - 8x - 2y + 1 = 0$ and $x^2 + y^2 + 6x + y = 0$ is:

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