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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
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B
56
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C
56
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D
34
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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).
$$\therefore          -40 - 16 - a - b = 0$$
$$\Rightarrow                     a+b = -56$$
109225_117083_ans_5cb711071b3d46dfa4f289e8fd8ed16d.png

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