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(C) The given circles are:$S_1: x^2+y^2-4x+6y-3=0$$S_2: x^2+y^2+2x-2y-2=0$Rewriting the equations in standard form $(x-h)^2+(y-k)^2=r^2$:$S_1: (x-2)^2

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$\frac{\sqrt{231}}{5}$

Vedclass · Answer

(C) The given circles are:$S_1: x^2+y^2-4x+6y-3=0$$S_2: x^2+y^2+2x-2y-2=0$Rewriting the equations in standard form $(x-h)^2+(y-k)^2=r^2$:$S_1: (x-2)^2+(y+3)^2 = 3+4+9 = 16 = 4^2$. Center $C_1 = (2, -3)$,radius $r_1 = 4$.$S_2: (x+1)^2+(y-1)^2 = 2+1+1 = 4 = 2^2$. Center $C_2 = (-1, 1)$,radius $r_2 = 2$.The distance between the centers $d = C_1C_2 = \sqrt{(2 - (-1))^2 + (-3 - 1)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9+16} = 5$.The common chord $AB$ is perpendicular to the line joining the centers. Let the common chord intersect $C_1C_2$ at $M$. Let $AM = h$ and $C_1M = x$. Then $C_2M = 5-x$.In $	riangle AC_1M$,$h^2 + x^2 = r_1^2 = 4^2 = 16$.In $	riangle AC_2M$,$h^2 + (5-x)^2 = r_2^2 = 2^2 = 4$.Subtracting the equations: $x^2 - (5-x)^2 = 16 - 4 = 12$.$x^2 - (25 - 10x + x^2) = 12$ $\Rightarrow 10x - 25 = 12$ $\Rightarrow 10x = 37$ $\Rightarrow x = 3.7$.$h^2 = 16 - (3.7)^2 = 16 - 13.69 = 2.31 = \frac{231}{100}$.$h = \sqrt{\frac{231}{100}} = \frac{\sqrt{231}}{10}$.The length of the common chord $AB = 2h = 2 	imes \frac{\sqrt{231}}{10} = \frac{\sqrt{231}}{5}$.

If $A$ and $B$ are the points of intersection of the circles $x^2+y^2-4x+6y-3=0$ and $x^2+y^2+2x-2y-2=0$,then the distance between $A$ and $B$ is

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