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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=` (A) 50 (B) 56 (C) `-56` (D) `-34`A. 50B. 56C. `-56`D. `-34` |
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Answer» Correct Answer - 3 The equation of radical axis ( i.e., common chord ) of the two circles is `10x+4ya-a-b=0` (1) The center of the first circle is `H(-4,-4)` Since the second circle bisects the circumference of the first circle, the center `H(-54,-4)` of the first circle must lie on the coomon chord (1). Therefore, ltbr. `-40-16-1-b=0` or `a+b= -56` |
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