The equation of the transverse common tangent | vedclass

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Question

(A) For circle $S_1: x^2+y^2-6x-8y+9=0$,the center is $C_1(3, 4)$ and radius $r_1 = \sqrt{3^2+4^2-9} = \sqrt{16} = 4$.For circle $S_2: x^2+y^2+2x-2y+1

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$4x+3y-4=0$

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(A) For circle $S_1: x^2+y^2-6x-8y+9=0$,the center is $C_1(3, 4)$ and radius $r_1 = \sqrt{3^2+4^2-9} = \sqrt{16} = 4$.For circle $S_2: x^2+y^2+2x-2y+1=0$,the center is $C_2(-1, 1)$ and radius $r_2 = \sqrt{(-1)^2+1^2-1} = \sqrt{1} = 1$.The distance between centers $C_1C_2 = \sqrt{(3 - (-1))^2 + (4 - 1)^2} = \sqrt{4^2 + 3^2} = 5$.Since $C_1C_2 = r_1 + r_2 = 4 + 1 = 5$,the circles touch each other externally.The transverse common tangent at the point of contact is the radical axis of the two circles,given by $S_1 - S_2 = 0$.$(x^2+y^2-6x-8y+9) - (x^2+y^2+2x-2y+1) = 0$.$-8x - 6y + 8 = 0$.Dividing by $-2$,we get $4x + 3y - 4 = 0$.

The equation of the transverse common tangent of the circles $x^2+y^2-6x-8y+9=0$ and $x^2+y^2+2x-2y+1=0$ is

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