The common tangent to the circles x^2 + y^2 = | vedclass

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(B) For the circle $x^2 + y^2 = 4$,the center is $C_1(0, 0)$ and the radius is $r_1 = 2$.For the circle $x^2 + y^2 + 6x + 8y - 24 = 0$,the center is $

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$(6, -2)$

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(B) For the circle $x^2 + y^2 = 4$,the center is $C_1(0, 0)$ and the radius is $r_1 = 2$.For the circle $x^2 + y^2 + 6x + 8y - 24 = 0$,the center is $C_2(-3, -4)$ and the radius is $r_2 = \sqrt{(-3)^2 + (-4)^2 - (-24)} = \sqrt{9 + 16 + 24} = \sqrt{49} = 7$.The distance between the centers is $d = \sqrt{(-3-0)^2 + (-4-0)^2} = \sqrt{9 + 16} = 5$.Since $d = |r_2 - r_1| = |7 - 2| = 5$,the circles touch each other internally.The equation of the common tangent at the point of contact is given by $S_1 - S_2 = 0$.$(x^2 + y^2 - 4) - (x^2 + y^2 + 6x + 8y - 24) = 0$$-6x - 8y + 20 = 0$$3x + 4y - 10 = 0$.Checking the options,the point $(6, -2)$ satisfies the equation: $3(6) + 4(-2) - 10 = 18 - 8 - 10 = 0$.

The common tangent to the circles $x^2 + y^2 = 4$ and $x^2 + y^2 + 6x + 8y - 24 = 0$ also passes through the point

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