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

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(B) Given the circle $S_1: x^2+y^2+8x-4y+c=0$ with center $C_1=(-4, 2)$ and radius $r_1=\sqrt{(-4)^2+2^2-c}=\sqrt{20-c}$.Given the circle $S_2: x^2+y^

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-$59$

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(B) Given the circle $S_1: x^2+y^2+8x-4y+c=0$ with center $C_1=(-4, 2)$ and radius $r_1=\sqrt{(-4)^2+2^2-c}=\sqrt{20-c}$.Given the circle $S_2: x^2+y^2+2x+4y-11=0$ with center $C_2=(-1, -2)$ and radius $r_2=\sqrt{(-1)^2+(-2)^2-(-11)}=\sqrt{1+4+11}=4$.Since the circles touch externally,the distance between centers $C_1C_2 = r_1+r_2$.$C_1C_2 = \sqrt{(-4 - (-1))^2 + (2 - (-2))^2} = \sqrt{(-3)^2 + 4^2} = \sqrt{9+16} = 5$.Thus,$5 = \sqrt{20-c} + 4$ $\Rightarrow \sqrt{20-c} = 1$ $\Rightarrow 20-c = 1$ $\Rightarrow c = 19$.Now,the circle $S_1$ cuts the circle $S_3: x^2+y^2-6x+8y+k=0$ orthogonally. The condition for orthogonality is $2g_1g_3 + 2f_1f_3 = c_1+c_3$.For $S_1$,$g_1=4, f_1=-2, c_1=c=19$.For $S_3$,$g_3=-3, f_3=4, c_3=k$.$2(4)(-3) + 2(-2)(4) = 19 + k$.$-24 - 16 = 19 + k$.$-40 = 19 + k$.$k = -59$.

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 $k$ is equal to

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