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Question

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 $$k=$$


A
59
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B
59
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C
19
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D
19
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Solution

The correct option is B $$-59$$
Let $$s_1 : x^2+y^2+8x-4y+c=0$$
$$\therefore$$ centre is $$4:(-4, 2)$$ and radius $$=\sqrt{20-c}$$
 and $$s_2: x^2+y^2+2x+4y-11=0$$
centre is $$C_2:(-1, -2)$$ and $$r_2=\sqrt{5+11}=4$$
$$\because$$ these two circle one touching each other
$$c_1c_2=r_1+r_2$$
$$\sqrt{9+16}=\sqrt{20-c}+4$$
$$\Rightarrow 5-4=\sqrt{20-c}$$
$$\Rightarrow c=19$$
$$s_3: x^2+y^2-6x+8y+k=0$$
$$\therefore$$ centre $$c_3:(3, -4) r_2=\sqrt{25-k}$$
Since, $$s_1$$ and $$s_3$$ cuts orthogonally,
$$r_1^2+r^2_3=(c_1c_3)^2$$
Hence, $$(20-c)+(25-k)=(\sqrt{45+36})^2$$
$$\Rightarrow k=26-85=-59$$

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