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$If the circumference of the circle x^{2} + y^{2} + 8 x + 8 y - b = 0 is bisected by the circle$ $x^{2} + y^{2} - 2 x + 4 y + a = 0 then a + b =$

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detailed solution

Correct option is C

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=

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$If the circumference of the circle x^{2} + y^{2} + 8 x + 8 y - b = 0 is bisected by the circle$ $x^{2} + y^{2} - 2 x + 4 y + a = 0 then a + b =$