Q.

If the circumference of the circle x2+y2+8x+8y−b=0 is bisected by the circle x2+y2−2x+4y+a=0 then a+b=

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a

-56

b

58

c

-58

d

54

answer is C.

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Detailed Solution

Common chord is 10x+4y−b−a=0 is passes through (−4,−4)⇒−40−16=a+b⇒a+b=−56
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If the circumference of the circle x2+y2+8x+8y−b=0 is bisected by the circle x2+y2−2x+4y+a=0 then a+b=