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
B
59
C
19
D
19

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, $$(20-c)+(25-k)=(\sqrt{45+36})^2$$$$\Rightarrow k=26-85=-59$$Maths

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