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If the circle x2+y2+4x+22y+l=0 bisects the circumference of the circle x2+y2−2x+8y−m=0,then l*m is equal to

(A) 25

(B) 625

(C) 125

(D) 425

I don't know the condition when one circle intersects the circumference of other circle. So could not solve this question. Can someone help me in this question?

Swapnil Milind Inamdar
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  • Are you sure +8y shouldn't be +4y in the second equation? – Phil H May 08 '18 at 15:13
  • Please use MathJax to format your mathematical expressions. You can find a tutorial and quick reference here. – amd May 08 '18 at 18:07
  • There’s something wrong with your problem statement. If you try a few values of $l$ and $m$ that satisfy the conditions, you’ll find that their product isn’t constant. Their sum is, but this constant sum isn’t one of the given choices. – amd May 08 '18 at 18:44

2 Answers2

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The first circle intersects the second circle at two points, and the arc length of the second circle between those two points is half the circumference of the second circle. This means that the line segment between the two points is a diameter, and includes the center of the circle.

Acccumulation
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I'm assuming there is something wrong with the second equation and if corrected to give an $r^2$ value of 5, then answer (B) looks to be the answer except these circles intercept at the origin $(0, 0)$ and $(3,-1)$.

Phil H
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  • More likely there’s something wrong elsewhere in the problem. The product $lm$ isn’t constant over all values of $l$ and $m$ that satisfy the conditions. Their sum is, but even that value isn’t one of the choices. – amd May 08 '18 at 18:45