The number of common tangents to the circles | vedclass

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(C) Given equations of the circles are:$S_1: x^2+y^2+4x-6y-12=0$$S_2: x^2+y^2-8x+10y+5=0$For $S_1$,the centre $C_1 = (-2, 3)$ and the radius $r_1 = \s

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Given equations of the circles are:$S_1: x^2+y^2+4x-6y-12=0$$S_2: x^2+y^2-8x+10y+5=0$For $S_1$,the centre $C_1 = (-2, 3)$ and the radius $r_1 = \sqrt{(-2)^2 + 3^2 - (-12)} = \sqrt{4+9+12} = \sqrt{25} = 5$.For $S_2$,the centre $C_2 = (4, -5)$ and the radius $r_2 = \sqrt{4^2 + (-5)^2 - 5} = \sqrt{16+25-5} = \sqrt{36} = 6$.The distance between the centres $C_1$ and $C_2$ is:$C_1C_2 = \sqrt{(4 - (-2))^2 + (-5 - 3)^2} = \sqrt{6^2 + (-8)^2} = \sqrt{36+64} = \sqrt{100} = 10$.Now,compare the distance $C_1C_2$ with the sum and difference of the radii:$r_1 + r_2 = 5 + 6 = 11$$|r_1 - r_2| = |5 - 6| = 1$Since $|r_1 - r_2| Therefore,the number of common tangents to the two circles is $2$.Hence,option $C$ is correct.

The number of common tangents to the circles $x^2+y^2+4x-6y-12=0$ and $x^2+y^2-8x+10y+5=0$ is

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