Question

# If the circumference of the circle $$x^2 + y^2 + 8x + 8y - b = 0$$ is bisected by the circle $$x^2 + y^2 - 2x + 4y + a = 0$$, then a+b equals to

A
50
B
56
C
56
D
34

Solution

## The correct option is B $$-56$$Equation of radical axis (i.e. common chord) of the two circles is$$10x + 4y - a - b = 0$$      ...... (i)Centre of first circle is $$H(-4, -4)$$Since second circle bisects the circumference of the first circle, therefore, centre $$H(-4, -4)$$ of the first circle must lie on the common chord Equation (i).$$\therefore -40 - 16 - a - b = 0$$$$\Rightarrow a+b = -56$$Maths

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