If the circle x^2+y^2+8x-4y+c=0 touches the c | vedclass

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(A) For the first circle $S_1: x^2+y^2+8x-4y+c=0$,the center is $C_1(-4, 2)$ and radius $r_1 = \sqrt{(-4)^2 + 2^2 - c} = \sqrt{20-c}$.For the second c

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-$59$

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(A) For the first circle $S_1: x^2+y^2+8x-4y+c=0$,the center is $C_1(-4, 2)$ and radius $r_1 = \sqrt{(-4)^2 + 2^2 - c} = \sqrt{20-c}$.For the second circle $S_2: x^2+y^2+2x+4y-11=0$,the center is $C_2(-1, -2)$ and radius $r_2 = \sqrt{(-1)^2 + (-2)^2 - (-11)} = \sqrt{1+4+11} = 4$.The distance between centers $C_1$ and $C_2$ is $d = \sqrt{(-4 - (-1))^2 + (2 - (-2))^2} = \sqrt{(-3)^2 + 4^2} = \sqrt{9+16} = 5$.Since the circles touch externally,$d = r_1 + r_2$,so $5 = \sqrt{20-c} + 4$,which implies $\sqrt{20-c} = 1$,so $20-c = 1$,giving $c = 19$.The third circle is $S_3: x^2+y^2-6x+8y+k=0$.Two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ cut orthogonally if $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.For $S_1$ and $S_3$,we have $g_1=4, f_1=-2, c_1=19$ and $g_3=-3, f_3=4, c_3=k$.Thus,$2(4)(-3) + 2(-2)(4) = 19 + k$.$-24 - 16 = 19 + k$,which gives $-40 = 19 + k$,so $k = -59$.

If the circle $x^2+y^2+8x-4y+c=0$ touches the circle $x^2+y^2+2x+4y-11=0$ externally and cuts the circle $x^2+y^2-6x+8y+k=0$ orthogonally,then $k=$

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