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If the circle \mathrm{x^2+y^2+4 x+22 y+1=0} bisects the circumference of the circle \mathrm{x^2+y^2-2 x+8 y-m=0}, then \mathrm{I+m} is equal to
 

Option: 1

60


Option: 2

50


Option: 3

40


Option: 4

56


Answers (1)

best_answer

The common chord of the given circles is

\mathrm{6 \mathrm{x}+14 \mathrm{y}+1+\mathrm{m}=0}                (1)

Since \mathrm{x^2+y^2+4 x+22 y+1=0} biscets the circumference of the circle

\mathrm{x^2+y^2-2 x+8 y-m=0}, (1) paases through the centre of the second circle i.e. \mathrm{( 1 , -4).}

Therefore, \mathrm{6-56+1+m=0 \Rightarrow 1+m=50.}

Hence option 2 is correct.

Posted by

jitender.kumar

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