The number of common tangents to the circles | vedclass

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(C) For the circle $x^2+y^2+2x+8y-23=0$,the center $C_1 = (-1, -4)$ and radius $r_1 = \sqrt{(-1)^2 + (-4)^2 - (-23)} = \sqrt{1 + 16 + 23} = \sqrt{40}

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) For the circle $x^2+y^2+2x+8y-23=0$,the center $C_1 = (-1, -4)$ and radius $r_1 = \sqrt{(-1)^2 + (-4)^2 - (-23)} = \sqrt{1 + 16 + 23} = \sqrt{40} = 2\sqrt{10}$.For the circle $x^2+y^2-4x-10y+19=0$,the center $C_2 = (2, 5)$ and radius $r_2 = \sqrt{(2)^2 + (5)^2 - 19} = \sqrt{4 + 25 - 19} = \sqrt{10}$.The distance between the centers $d = C_1C_2 = \sqrt{(2 - (-1))^2 + (5 - (-4))^2} = \sqrt{3^2 + 9^2} = \sqrt{9 + 81} = \sqrt{90} = 3\sqrt{10}$.Since $r_1 + r_2 = 2\sqrt{10} + \sqrt{10} = 3\sqrt{10}$,we have $d = r_1 + r_2$.Because the distance between the centers is equal to the sum of the radii,the two circles touch each other externally.When two circles touch each other externally,they have exactly $3$ common tangents.

The number of common tangents to the circles $x^2+y^2+2x+8y-23=0$ and $x^2+y^2-4x-10y+19=0$ is:

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