The radical axis of the circles S_1: x^2+y^2- | vedclass

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(A) The radical axis of two circles $S_1=0$ and $S_2=0$ is given by $S_1 - S_2 = 0$. $ (x^2+y^2-4x+6y-10) - (x^2+y^2+2x-6y+2) = 0 $ $ -6x + 12y - 12 =

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two real and distinct points

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(A) The radical axis of two circles $S_1=0$ and $S_2=0$ is given by $S_1 - S_2 = 0$. $ (x^2+y^2-4x+6y-10) - (x^2+y^2+2x-6y+2) = 0 $ $ -6x + 12y - 12 = 0 $ $ x - 2y + 2 = 0 $ This is the equation of the radical axis. To find the intersection with $S_1$,we substitute $x = 2y - 2$ into $S_1$: $ (2y-2)^2 + y^2 - 4(2y-2) + 6y - 10 = 0 $ $ 4y^2 - 8y + 4 + y^2 - 8y + 8 + 6y - 10 = 0 $ $ 5y^2 - 10y + 2 = 0 $ The discriminant $D = b^2 - 4ac = (-10)^2 - 4(5)(2) = 100 - 40 = 60$. Since $D > 0$,the quadratic equation has two real and distinct roots for $y$,which implies the radical axis cuts the circle $S_1$ at two real and distinct points.

The radical axis of the circles $S_1: x^2+y^2-4x+6y-10=0$ and $S_2: x^2+y^2+2x-6y+2=0$ cuts the circle $S_1$ in

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