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

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

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-$59$

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(B) Given the circle $x^2+y^2+8x-4y+c=0$,its center $C_1 = (-4, 2)$ and radius $r_1 = \sqrt{(-4)^2 + 2^2 - c} = \sqrt{20-c}$.For the circle $x^2+y^2+2x+4y-11=0$,its 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,$C_1C_2 = r_1 + r_2$.$C_1C_2 = \sqrt{(-4 - (-1))^2 + (2 - (-2))^2} = \sqrt{(-3)^2 + 4^2} = \sqrt{9+16} = 5$.So,$5 = \sqrt{20-c} + 4$,which implies $\sqrt{20-c} = 1$,so $20-c = 1$,hence $c = 19$.Now,the circle $x^2+y^2+8x-4y+19=0$ cuts $x^2+y^2-6x+8y+k=0$ orthogonally.The condition for orthogonality is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.Here,$g_1 = 4, f_1 = -2, c_1 = 19$ and $g_2 = -3, f_2 = 4, c_2 = k$.$2(4)(-3) + 2(-2)(4) = 19 + k$.$-24 - 16 = 19 + k$.$-40 = 19 + k$.$k = -59$.

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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